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Principles of Mathematical Analysis 2

Paolo Guiotto

Contents

Chapter 1. Basic Differential Equations 11.1. Differential Equations in Applications 11.2. First order linear equations 41.3. First order separable variables equations 81.4. Second Order Linear Equations 111.5. Exercises 17

Chapter 2. Topology in Rd 212.1. Euclidean norm 212.2. Limit 242.3. Continuity 292.4. Properties of continuous functions 302.5. Exercises 32

Chapter 3. Differential Calculus 353.1. Directional derivative 353.2. Differential 373.3. Extrema 403.4. Exercises 43

Chapter 4. Constrained Optimization 454.1. Implicit functions: scalar case 464.2. Lagrange multipliers: scalar case 484.3. Dini Theorem: case of systems 504.4. Lagrange Multipliers: general case 524.5. Exercises 57

Chapter 5. Integration 595.1. Integral 605.2. Reduction formula 635.3. Change of variables 685.4. Barycenter, center of mass, inertia moments 735.5. Exercises 75

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CHAPTER 1

Basic Differential Equations

Ordinary Differential Equations (ODEs) is a wide topic of Mathematical Analysis. Their relevanceis due to applications in Physics, Engineering, Biology, Economy, etc. An ODE is first of all an equation,that is an identity with an unknown. Differently from algebraic equations, the unknown is a function ofone variable, let say y = y(t) (t ∈ R). The equation expresses a relation between y(t) and some of itsderivatives y′(t), y′′(t), . . . up to a certain maximum order, called order of the equation. This explainsthe D and the E of ODE. The O is to distinguish these equations to similar equations but with unknownfunction depending by several variables (this type of equations is called Partial Differential Equations,PDEs).

In this Chapter we introduce to the simplest type of ODEs: first and second order linear equations, firstorder separable variables equations. To understand importance of these equations, we will accompanytheory with several applied examples.

1.1. Differential Equations in Applications

In this Section we show how ODEs arise in applied problems. Reader should focus not only to themathematical side of the story, rather on the modelling process leading to an ODE.

Example 1.1.1. A water vessel looses, by percolation, a fraction of the volume of water therein containedat constant time rate ν. The vessel is refilled by a constant flux φ. Discuss the time behaviour of thevolume of water contained the vessel. In particular, what happens in long time? Does the volume ofwater reach some stable level?Sol. — Let’s call V(t) the volume of the water inside the vessel, Vmax the maximum volume and V(0) the initialvolume. The variation of volume V(t) on a small time interval, that is V(t + dt) − V(t) is given by a reduction−νV(t) dt due to percolation and by an increase φ dt due to the flux. Therefore

V(t + dt) − V(t) = −νV(t) dt + φ dt.

Dividing by dt and letting dt −→ 0 we get the equation

V ′(t) = −νV(t) + φ.

Leaving apart the formal details, we could notice that the equation is equivalent toV ′(t)

−νV(t) + φ= 1.

At the left hand side we recognize, more or less, a derivative. Indeed

(log | − νV(t) + φ|)′ =−νV ′(t)−νV(t) + φ

= −ν,

1

2

by the equation. So we could say thatlog | − νV(t) + φ| = −νt + C,

where C is some constant that can be found imposing that V(0) = V0 that is C = log | − νV0 + φ|. Now, we wouldhave

| − νV(t) + φ| = e−νt+C, ⇐⇒ V(t) =φ

ν±

1ν

e−νt+C,

But + or −? According the value for C, V0 =φν ±

1ν | − νV0 + φ| =

φν ±

��φν − V0

��. We see that if φν > V0 we have totake − otherwise +. Therefore

V(t) =

φν −

φ−νV0ν e−νt = φ

ν

(1 − e−νt

)+ V0e−νt, if φ

ν > V0,

φν +

νV0−φν e−νt = φ

ν

(1 − e−νt

)+ V0e−νt, if φ

ν < V0.

=φ

ν

(1 − e−νt

)+ V0e−νt .

As time goes on, V stabilizes through a limit level V(+∞) = φν .

The equation V ′(t) = −νV(t) + φ is an example of first order linear equation. Roughly, linear means thatthe dependence on V and V ′ is through a first degree polynomial. As we will see for ODE this means aparticularly simple structure for solutions of the equations and explicit formulas for the solution.

Example 1.1.2 (Newton’s Equations). The most classical example of ODE is given by Newton’sequations, direct consequence of Newton’s second law. A particle of mass m in movement under theeffect of some force ®F fulfils,

m®a = ®F,where ®a is the acceleration of the particle. For sake of simplicity, we assume the mass moving on a straightrail, we can characterize its position in terms of a function x = x(t), t representing time. Accelerationis then x ′′(t). As for acceleration, force ®F can be identified with a scalar F. In general, Physical forcesdepend by position x(t) (as in the case of gravitational forces or elastic forces), velocity x ′(t) (as in thecase of friction) or directly by time t (if intensity of applied force is changing in time). Therefore, Newton’ssecond law assumes the form

mx ′′(t) = F(t, x(t), x ′(t)).A classical example is that one of a mass m moving under the action of a elastic force and subject to friction. Ifκ > 0 is the elastic constant and assuming the origin as the rest position, first force is

−κx(t).

The minus means that when the particle is out of the origin the force tends to move the particle toward the origin.Second component of applied force is friction, which depends on velocity. For simplicity, we will assume the railbe homogeneous and friction be proportional to velocity in a way to decelerate the mass. This means the force is

−νx ′(t),

(ν > 0 is called viscosity). Putting all together we obtain the equationmx ′′(t) = −κx(t) − νx ′(t).

If the viscosity vanishes (harmonic oscillator), we expect a perfect infinite oscillatory motion, that is x(t) like asinusoid, whereas if we set off the spring and we leave only the viscosity we expect that if the mass has some initialvelocity, it will tends to stop sometime later. Combining the two we expect an attenuated oscillation motion. But:how to prove it? And what about if we would have a precise measure of the attenuation?

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If an external force f (t) (that is, independent by the mass) acts on the mass the equation, the equation assumesthe form

mx ′′(t) = −κx(t) − νx ′(t) + f (t).With this simple equation we describe lots of phenomena like forced oscillations. An interesting and surprisingexample is the case of resonance. Imagine a periodic external force is applied to an harmonic oscillator,

mx ′′(t) = −κx(t) + F0 sin(ωt).

It turns out that if ω =√κ external force enters in resonance with elastic force leading to an x with oscillation

amplitude increasing in time. A model like this was used to provide a simple explanation of the famous Takomabridge collapse.

This is a second order linear equation, linear because the dependence by x, x ′, x ′′ is by a so called linearfunction, that is a first degree polynomial. In the two previous Examples, the unknown function had timeas independent variable. Of course this is not the unique possible parametrization.

Example 1.1.3 (Catenary Problem). A chain is suspended by two fixed points: what is the curve thehanging chain assumes under its own weight when supported only at its ends? In his Two New Sciences(1638), Galileo says that a hanging cord is an approximate parabola. But what precisely is this curve?Sol. — Let xy be the plane containing the curve, we use x as parameter in such a way that α = α(x) is the ordinatecorresponding to x on the catenary. Our goal is to determine the function α. Let’s see how the Mechanics of theproblem enters to determine a Differential Equation for α.

Consider a small portion of the catenary included between points (x, α(x)) and (x + ∆x, α(x + ∆x)) (∆x > 0be "small"). On this portion of catenary the following forces act: the tension exercised at two extremities bythe remaining parts of the catenary and the gravitation. This last is easy because pulls downward as m®g. Here®g = (0,−g) (g = 9.8m/s2) while m is the mass of the small portion of catenary. Let’s say that % is mass lineardensity, m = % · ds where ds =length of the portion. By Pythagorean Thm

ds ∼√(∆x)2 + (α(x + ∆x) − α(x))2,

the approximation being precise as ∆x ≈ 0. Because this is the case we’re considering, α(x + ∆x) − α(x) =α′(x)∆x + o(∆x) ∼ α′(x)∆x, whence

m = %√

1 + α′(x)2∆x.In conclusion

m®g =(0,−%g

√1 + α′(x)2∆x

).

About the tension, let’s denote by ®T(y) the force exercises by the part of the catenary included between (y, α(y))and B (final point). Therefore, the force exercised by the part of the catenary included between A and (y, α(y))must

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be −®T(y). Therefore, in (x, α(x)) is acting −®T(x), in (x + ∆x, α(x + ∆x)) is acting ®T(x + ∆x) and these two are inequilibrium with m®g, that is

−®T(x) + ®T(x + ∆x) + m®g = ®0,or, in components ®T = (τ,σ),

τ(x + ∆x) − τ(x) = 0,

σ(x + ∆x) − σ(x) − %g√

1 + α′(x)2∆x = 0.The first one says that τ(x) ≡ τ0 is constant. The second one, dividing by ∆x and letting this to 0, says

σ′(x) = %g√

1 + α′(x)2.

There’s still one more information we need to use: of course ®T is tangent to the catenary at each of its points. Now,because ®T = (τ0, σ(x)) we need that

σ(x)τ0= α′(x), σ(x) = τ0α

′(x).

By this we obtain finally σ′(x) = τ0α′′(x), thus

(1.1.1) α′′(x) =%g

τ0

√1 + α′(x)2.

This is the catenary equation and it is a non linear equation.

1.2. First order linear equations

The first type of ODE we consider is the following

(1.2.1) y′(t) = a(t)y(t) + b(t), t ∈ I .

Here a, b : I ⊂ R −→ R are known function (called coefficients). If b ≡ 0 we say that the equation ishomogeneous. In this case the set of solutions has a linear structure. Indeed we can notice that if ϕ andψ are solutions, then also αϕ + βψ is a solution (here α, β ∈ R). Indeed

(αϕ + βψ)′(t) = αϕ′(t) + βψ ′(t) = αa(t)ϕ(t) + βa(t)ψ(t) = a(t)(αϕ + βψ)(t), t ∈ I .

Homogeneous equations are simpler. Roughly,

y′(t) = a(t)y(t)if y,0⇐⇒

y′(t)y(t)

= a(t), ⇐⇒ (log |y(t)|)′ = a(t),

thus, at least if y , 0,

log |y(t)| =∫

a(t) dt + c,

where c is a constant. Therefore

|y(t)| = ece∫a(t) dt, ⇐⇒ y(t) = ±ece

∫a(t) dt ≡ ce

∫a(t) dt, c ∈ R.

Notice that this formula produces, for c = 0, y ≡ 0 which is a solution for the homogeneous equation.Formula we just obtained could work for y , 0 and y ≡ 0 solutions. But who says that these are all thepossible solutions? Rather than fixing this argument, we provide a simpler but less "intuitive" proof thatshows this characterization holds true:

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Proposition 1.2.1. Let a ∈ C (I), I ⊂ R interval. All the solutions of the homogeneous equationy′(t) = a(t)y(t), t ∈ I,

are(1.2.2) y(t) = ce

∫a(t) dt, c ∈ R.

Proof — Define A(t) =∫

a(t) dt. Then(e−Ay

) ′= −A′e−Ay + e−Ay′ = e−A (−ay + y) .

Now, because y is a solution iff y′ − ay = 0, we have also,

y solution ⇐⇒(e−Ay

) ′≡ 0,

and because I is an interval, we deduce

y solution ⇐⇒ e−Ay ≡ c, ⇐⇒ y(t) = ce−A(t) = ce−∫a(t) dt .

Let’s move to the general case of a non homogeneous equation,y′(t) = a(t)y(t) + b(t).

We prove now that the general solution is obtained by summing to (1.2.2) a particular solution of the nonhomogeneous equation.

Proposition 1.2.2. Let a, b ∈ C (I), I ⊂ R interval. If U = U(t) is a particular solution of the nonhomogeneous equation, then the general solution of

y′(t) = a(t)y(t) + b(t), t ∈ I,

is(1.2.3) y(t) = ce

∫a(t) dt +U(t), t ∈ I .

Proof — Just notice that y is a solution of the non homogeneous equation iff

(y −U)′ = y′ −U ′ = (ay + b) − (aU + b) = a(y −U),(1.2.2)⇐⇒ y(t) −U(t) = ce

∫a(t) dt .

Thus, to determine the general solution for the non homogeneous equation it remains to determine aparticular solution. This may be determined through the so called method of variation of constants:

Theorem 1.2.3. Let a, b ∈ C (I), I ⊂ R interval. Then, a particular solution fory′(t) = a(t)y(t) + b(t), t ∈ I,

is

(1.2.4) U(t) = e∫a(t)dt

∫e−

∫a(t)dtb(t) dt, t ∈ I .

Thus, the general solution of the non homogeneous equation is

(1.2.5) y(t) = e∫a(t)dt

[∫e−

∫a(t)dtb(t) dt + c

], t ∈ I,

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where c ∈ R is a constant. The (1.2.5) is also called general integral.Proof — We start by searching for U = U(t) particular solution. The idea is to look at

U(t) = c(t)e∫a(t) dt, where c = c(t) has to be determined by imposing U ′ = aU + b.

Now, being

U ′(t) =(c(t)e

∫a(t) dt

) ′= c′(t)e

∫a(t) dt + c(t)e

∫a(t) dta(t) = e

∫a(t) dt (c′(t) + a(t)c(t)) ,

we haveU ′ = aU + b, ⇐⇒ e

∫a(t) dt (c′(t) + a(t)c(t)) = a(t)c(t)e

∫a(t) dt + b(t),

that is

c′(t)e∫a(t) dt = b(t), ⇐⇒ c′(t) = e−

∫a(t) dtb(t), ⇐⇒ c(t) =

∫e−

∫a(t) dtb(t) dt + c, c ∈ R.

Because any of these c(t) is good, we may take c = 0, this leading to (1.2.4). Together with (1.2.3), finally weobtain

y(t) = ce∫a(t) dt +

(∫e−

∫a(t) dtb(t) dt

)e∫a(t) dt = e

∫a(t) dt

(∫e−

∫a(t) dtb(t) dt + c

),

which is (1.2.5).

Example 1.2.4. Find the general integral for the equation

y′(t) −2ty(t) = 1, t ∈]0,+∞[.

Sol. — We havey′(t) =

2ty(t) + 1 = a(t)y(t) + b(t), where a(t) =

2t, b(t) = 1.

Therefore

y(t) = e∫ 2

t dt

(∫e−

∫ 2t dtdt + c

)= e2 log t

(∫e−2 log tdt + c

)= t2

(∫1t2 dt + c

)= t2

(−

1t+ c

)= −t + ct2.

You shouldn’t be surprised because uniqueness does not hold for ODEs. Just the simplest of equations

y′ = 0,

has infinitely many solutions (all constants). However, further conditions may lead to a unique solution.A very important case is the so called Cauchy Problem or passage problem or, again, initial conditionproblem. This consists in finding a solution of an ODE fulfilling a passage/initial value condition.Formally, this problem may be stated in the following form:

(1.2.6) CP(t0, y0)

y′(t) = a(t)y(t) + b(t), t ∈ I,

y(t0) = y0.

Here, of course, t0 ∈ I. It is easy to check that this problem has a unique solution:

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Corollary 1.2.5. Let a, b ∈ C (I), I ⊂ R interval. For every t0 ∈ I, Cauchy Problem CP(t0, y0) admitsa unique solution.Proof — Because general solution has the form

y(t) = ce∫a(t) dt +U(t) ≡ ceA(t) +U(t), where A(t) =

∫a(t) dt,

being U the particular solution, we have that y solves CP(t0, y0) iff

y0 = ceA(t0) +U(t0), ⇐⇒ c =y0 −U(t0)

e−A(t0).

This c clearly exists (e−A(t0) , 0) and it is unique, thus we have existence and uniqueness for CP(t0, y0).

Example 1.2.6. Solve the Cauchy Problemy′(t) −

2y(t)1 − t2 = t, t > 1

y(2) = 0.

Sol. — Rewriting the equation in the canonical form

y′(t) =2

1 − t2 y(t) + t, =⇒ y(t) = e∫ 2

1−t2dt

(∫e−

∫ 21−t2

dt t dt + C).

Now ∫2

1 − t2 dt =∫

2(1 − t)(1 + t)

dt =∫

11 − t

+1

1 + tdt = −

∫1

t − 1dt + log |1 + t | = log

�� t + 1t − 1

�� .Because t ∈]1,+∞[, t+1

t−1 > 0, therefore

y(t) = elog t+1t−1

(∫e− log t+1

t−1 t dt + C)=

t + 1t − 1

(∫t − 1t + 1

t dt + C).

Now∫tt − 1t + 1

dt =∫

tt + 1 − 2

t + 1dt =

∫t dt − 2

∫t

t + 1dt =

t2

2− 2

∫dt + 2

∫1

t + 1dt =

t2

2− 2t + 2 log |t + 1|,

and finally

ϕ(t) =t + 1t − 1

(t2

2− 2t + 2 log(1 + t) + C

), t ∈]0,+∞[.

Imposing ϕ(2) = 0 we have

2(2 − 4 + 2 log 3 + C) = 0, ⇐⇒ C = 2(1 − log 3).

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1.3. First order separable variables equations

An equation of typey′(t) = a(t) f (y(t))

is called separable variables equation. This type of equations is, in some sense, an extension of firstorder homogeneous equations

y′(t) = a(t)y(t), assuming f (y) := y.

This remark suggests a possible method to solve the equation, the so called method of separation ofvariables. To understand better the argument let’s consider the Cauchy problem

CP(t0, y0)

y′(t) = a(t) f (y(t)), t ∈ I,

y(t0) = y0.

Here we assumea ∈ C (I), f ∈ C (J), I, J ⊂ R, intervals.

In order CP(t0, y0) make sense, here we need t0 ∈ I and y0 ∈ J. Indeed, setting t = t0 in the equation wemust have

y′(t0) = a(t0) f (y(t0)) = a(t0) f (y0),

that is both a(t0) (thus t0 ∈ I) and f (y0) (thus y0 ∈ J) must be well defined. Now we have the followingalternative:

• if f (y0) = 0, then clearly y(t) ≡ y0 is a solution because

y′(t) = 0, a(t) f (y(t)) = a(t) f (y0) = 0.

In this case y(t) ≡ y0 (constant solution) is a solution of the equation.• if f (y0) , 0, then f (y(t0)) , 0 and, by continuity, f (y(t)) , 0 in some neighbourhood of t0. Insuch a neighbourhood, we may write

y′(t) = a(t) f (y(t)), ⇐⇒y′(t)

f (y(t))= a(t).

This step is called separation of variables because it consists in writing all the terms containingy on one side, leaving on the other an explicit expression of t. As in the case of first orderlinear homogeneous equations, we may see at lhs as the derivative of something. In other terms,taking side by side the primitives, we have∫

y′(t)f (y(t))

dt =∫

a(t) dt + c,

where c is an arbitrary constant. The rhs can be explicitly computed. Conversely, the lhscontains the unknown y. However, we may notice that∫

y′(t)f (y(t))

dtu:=y(t), du=y′(t) dt

=

∫1

f (u)du =: G(u)

��u=y(t)

= G(y(t)).

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Therefore, assuming G =∫

1f known, we have an algebraic equation for y,

(1.3.1) G(y(t)) =∫

a(t) dt + c.

This is called implicit form of the solution. If G is also invertible, we may go on to obtain

(1.3.2) y(t) = G−1(∫

a(t) dt + c).

This is the explicit form. The value of c is determined imposing the passage condition y(t0) = y0.Apparently, this argument shows that either y is constant (if f (y0) = 0) or y is given by (1.3.1) inimplicit form or (1.3.2) in explicit form. There’s, however, a subtle point: these two alternatives hold upf (y(t)) , 0. So the question is whether it is possible or less that f (y(t)) , 0 always. This turns out to betrue if we require something more on f :

Theorem 1.3.1. Let a ∈ C (I) and f ∈ C 1(J) with I, J ⊂ R intervals. Then, for any (t0, y0) ∈ I × J thereexists a solution y of

y′(t) = a(t) f (y(t)),

y(t0) = y0.

In particular• if f (y0) = 0 then y(t) ≡ y0 is a (stationary) solution.• if f (y0) , 0 then, setting G(z) :=

∫1

f (z) dz a primitive of 1/ f , we have

(1.3.3) G(y(t)) =∫

a(t) dt + c,

for a suitable c. If G is invertible we have

(1.3.4) y(t) = G−1(∫

a(t) dt + c).

The (1.3.3) is called implicit form for the solution, the (1.3.4) is called explicit form for thesolution.

In particular: existence and uniqueness for the Cauchy problem holds true.Proof — Omitted.

Example 1.3.2. Solve the Cauchy problemy′(t) = 1 + y(t)2,

y(0) = y0.

Sol. — This is a separable variables equation y′ = a(t) f (y) with a(t) ≡ 1 and f (y) = 1 + y2 clearly fulfills thehypotheses of the previous Thm. Of course because f (y0) = 1 + y2

0 > 0 we don’t have stationary solutions and we

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can separate variables:

y′(t) = 1 + y(t)2, ⇐⇒y′(t)

1 + y(t)2= 1, ⇐⇒

∫y′(t)

1 + y(t)2dt =

∫1 + C = t + C.

Now, ∫y′(t)

1 + y(t)2dt =

∫1

1 + u2 du��u=y(t)

= arctan u|u=y(t) = arctan y(t),

soarctan(y(t)) = t + C, ⇐⇒ y(t) = tan(t + C).

Imposing y(0) = y0 we get y0 = tan C, that is C = arctan y0 and finally

y(t) = tan(t + arctan y0).

Evidently this solution "lives" up to the time when t + arctan y0 =π2 (in the future) and t + arctan y0 = −

π2 (in the

past), that is in this case

It0 =]−π

2− arctan y0,

π

2− arctan y0

[.

Example 1.3.3 (Catenary). Solve the catenary equation (1.1.1).Sol. — Setting α′(x) =: y(x) we have the first order equation

y′ =%g

τ0

√1 + y2,

which is a particular case of separable variables equation. Here a(x) ≡ %gτ0, f (y) =

√1 + y2. Clearly a ∈ C (R)

and f ∈ C 1(R). Notice also that f , 0 always. Thus, to determine solutions we can separate variables,

y′√1 + y2

=%g

τ0, ⇐⇒

∫y′√

1 + y2dx =

%g

τ0x + γ.

Now, ∫y′√

1 + y2dt

u=y(x)=

∫1

√1 + u2

du = sinh−1 u = sinh−1 y(x).

In conclusion

sinh−1 y(x) =%g

τ0x + γ, ⇐⇒ y(x) = sinh

(%g

τ0x + γ

).

Finally, because y = α′,

α(x) =∫

sinh(%g

τ0x + γ

)dx + γ =

τ0%g

cosh(%g

τ0x + γ

)+ γ.

For instance assume that A = (−`/2, h), B = (`/2, h). Parameters γ, γ are determined by solvingτ0%g cosh

(− `2

%gτ0+ γ

)+ γ = h,

τ0%g cosh

(`2%gτ0+ γ

)+ γ = h.

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Taking the difference, cosh(− `2

%gτ0+ γ

)= cosh

(`2%gτ0+ γ

), and because cosh is even, the unique possibility if that

− `2%gτ0+ γ = −

(`2%gτ0+ γ

)that is, γ = 0. Therefore, γ = h − τ0

%g cosh `2%gτ0, whence

α(x) =τ0%g

cosh%g

τ0x + h −

τ0%g

cosh`

2%g

τ0.

1.4. Second Order Linear Equations

More involved the theory for second order linear ODE, that is for equations of type(1.4.1) y′′(t) = a(t)y′(t) + b(t)y(t) + f (t).

If f ≡ 0, the equation is said homogenous and if a(t) ≡ a, b(t) ≡ b the equation is said to have constantcoefficients. We will limit to this case for simplicity that, for convenience, we will rewrite as

y′′ + ay′ + by = f (t).

To begin we will consider the homogeneous casey′′(t) + ay′(t) + by(t) = 0.

To solve in general this equation notice the following: if we call D the derivative, then previous equationcan be rewritten as

(D2 + aD + b)y = 0.The polynomial

λ2 + aλ + bis called characteristic polynomial and basically it contains all the information to look for solutions.

Theorem 1.4.1. The general integral of y′′ + ay′ + by = 0 isc1w1(t) + c2w2(t), c1, c2 ∈ R,

where• if ∆ = a2 − 4b > 0, (w1,w2) = (eλ1t, eλ2t ) with λ1,2 are the roots of the char. pol.;• if ∆ = 0, (w1,w2) = (eλ1t, teλ1t ) with λ1 is the unique root of the char. pol.;• if ∆ < 0, (w1,w2) = (eαt cos(βt), eαt sin(βt)) with λ1,2 = α ± iβ are the complex roots of thechar. pol.

The couple (w1,w2) is called fundamental system of solutions.Proof —We develop the proof on three cases, according ∆ > 0, ∆ = 0, ∆ < 0.Case ∆ > 0: the characteristic polynomial can be factorized as

λ2 + aλ + b = (λ − λ1)(λ − λ2),

so we may expect thatD2 + aD + b = (D − λ1)(D − λ2),

hence(D2 + aD + b)y = 0, ⇐⇒ (D − λ1)(D − λ2)y = 0.

Now call ψ = (D − λ2)y. Then(D − λ1)ψ = 0, ⇐⇒ ψ ′ = λ1ψ, ⇐⇒ ψ = ceλ1t .

12

But then(D − λ2)y = c1eλ1t, ⇐⇒ y′ = λ2y + c1eλ1t .

This is a first order linear equation that may be easily solved by the general formula (1.2.5), obtaining

y(t) = eλ2t

(∫e−λ2tc1eλ1t dt + c2

)= eλ2t

(c1

∫e(λ1−λ2)t dt + c2

)=

c1λ1 − λ2

eλ1t + c2eλ2t,

and being c1, c2 arbitrary, we get finallyy(t) = c1eλ1t + c2eλ2t .

Case ∆ = 0: we can repeat the same computations as before, just to the point

y(t) = eλ2t

(c1

∫e(λ1−λ2)t dt + c2

),

but now λ1 = λ2, therefore

y(t) = eλ1t

(c1

∫dt + c2

)= c1teλ1t + c2eλ2t .

Case ∆ < 0: in this case the characteristic polynomial is irreducible. However

λ2 + aλ + b =(λ +

a2

)2+

4b − a2

4= (λ − α)2 + β2 = 0, ⇐⇒ (λ − α)2 = −β2.

Therefore(D2 + aD + b)y = 0, ⇐⇒ (D − α)2y = −β2y.

Now, notice that(D − α)y = y′ − αy = eαtD

(e−αt y

),

whence(D − α)2y = eαtD

(e−αt

(eαtD

(e−αt y

) ) )= eαtD2 (

e−αt y).

By this(D − α)2y = −β2y, ⇐⇒ eαtD2 (

e−αt y)= −β2y, ⇐⇒ D2 (

e−αt y)= −β2e−αt y.

Finally, setting for a while φ = e−αt y, the previous equation becomes

D2φ = −β2φ,

and two solutions of this are φ1(t) = cos(βt), φ2(t) = sin(βt). This means that

y1(t) = eαt cos(βt), y2(t) = eαt sin(βt)

are two solutions of the initial equation and the general integral is in this case

c1eαt cos(βt) + c2eαt sin(βt).

As for linear first order equations, the general solution can be obtained by the general solution of thehomogeneous equation by adding a particular solution:

Proposition 1.4.2. Let (w1,w2) be a fundamental system of solutions for the homogeneous equation

(1.4.2) y′′ + ay′ + by = 0,

and U a particular solution of the equation

(1.4.3) y′′ + ay′ + by = f (t).

13

Then, the general integral of (1.4.3) isy(t) = c1w1(t) + c2w2(t) +U(t), c1, c2 ∈ R.

Proof — Just notice that if y solves (1.4.3), then y −U solves (1.4.2). Indeed(y −U)′′ + a(y −U)′ + b(y −U) = y′′ + ay′ + by − (U ′′ + aU ′ + bU) = f − f = 0, =⇒ y −U = c1w1 + c2w2.

To complete the solution of the second order non homogeneous equation, we need to determine an itsparticular solution. As for the first order case, this may be determined through the method of variation ofconstants. This consists in looking to U of type

U(t) = c1(t)w1(t) + c2(t)w2(t), t ∈ I .

To find these coefficients, we impose that U be a solution of the equation.

Theorem 1.4.3 (Lagrange). Let (w1,w2) a fundamental system of solutions of (1.4.2). Define

W(t) := detw1 w2

w′1 w′2

,the wronskian of (w1,w2). Then, W(t) , 0 for all t and

(1.4.4) U(t) = −(∫

w2(t)W(t)

f (t) dt)w1(t) +

(∫w1(t)W(t)

f (t) dt)w2(t), t ∈ I,

is a particular solution of (1.4.3).Proof — Let U = c1w1 + c2w2 with cj ≡ cj(t) j = 1,2. Then

U ′ = c′1w1 + c1w′1 + c′2w2 + c2w

′2.

To simplify computations we impose the conditionc′1w1 + c′2w2 = 0.

ThenU ′′ = c′1w

′1 + c1w

′′1 + c′2w

′2 + c2w

′′2 .

HenceU ′′ = aU ′ + bU + f , ⇐⇒ c′1w

′1 + c′2w

′2 = f .

We may conclude that U is a solution iff

(1.4.5)

c′1w1 + c′2w2 = 0,

c′1w′1 + c′2w

′2 = f .

This can be seen as a 2 × 2 linear system in the unknown (c′1, c′2) and coefficients the matrix

w1 w2

w′1 w′2

.Now, to find c′1, c

′2 we apply the Cramer rule. Calling W(t) the determinant of the previous matrix,

W(t) := w1w′2 − w2w

′1, (wronskian of (w1,w2))

14

it is easy to check that in all cases W(t) , 0 for any t:

det

eλ1t eλ2t

λ1eλ1t λ2eλ2t

= e(λ1+λ2)t (λ2 − λ1). det

eλt teλt

λeλt (1 + λt)eλt

= e2λt,

and

det

eαt cos(βt) eαt sin(βt)

eαt (α cos(βt) − β sin(βt)) eαt (α sin(βt) + β cos(βt))

= βe2αt .

Therefore, by Cramer rule,

c′1(t) =−w2(t) f (t)

W(t), c′2(t) =

w1(t) f (t)W(t)

.

that is

(1.4.6) c1(t) = −∫

w2(t)W(t)

f (t) dt, c2(t) =∫

w1(t)W(t)

f (t) dt,

so we get just the (1.4.4).

Example 1.4.4. Find the general integral of the equation

y′′(t) + y′(t) − 6y(t) = 2e−t, t ∈ R.

Sol. —We start computing the fundamental system of solutions of the homogeneous equation. The characteristicpolynomial is

λ2 + λ − 6 = 0, ∆ = 1 + 24 = 25 > 0, λ± =−1 ±

√25

2=−1 ± 5

2= 2,−3.

Therefore the fundamental solutions are w1(t) = e2t , w2(t) = e−3t with wronskian

W(t) = (−3 − 2)e−t = −5e−t .

By Lagrange formula (1.4.4) we have

U(t) = −∫

e−3t

−5e−t2e−t dt e2t +

∫e2t

−5e−t2e−t dt e−3t =

25

∫e−3t dt e2t −

25

∫e2t dt e−3t

= −215

e−t −210

e−t = −13

e−t .

Therefore, the general integral is

y(t) = c1e2t + c2e−3t −13

e−t, c1, c2 ∈ R.

In the case of second order equations we see an interesting phenomenon: general integral depends ontwo free constants c1, c2. It is therefore clear that a unique condition on y is not sufficient to determineuniquely a solution. That’s why the Cauchy problem for a second order ODE is different respect to thefirst order case. Intuitively, we need a second condition on the solution. There’re two interesting cases:

15

• the Cauchy problem, that consists in finding a solution y fulfilling two initial conditions as

CP(t0, y0, y′0)

y′′ + ay′ + by = f (t),y(t0) = y0,y′(t0) = y′0.

• the boundary value problem, that consists in finding a solution y fulfilling two passageconditions as

BV(t0, y0; t1, y1)

y′′ + ay′ + by = f (t),y(t0) = y0,y(t1) = y1.

These are two entirely different problems. The first one has a strict interest in Physics, for instance. Here,the ODE corresponds to Newton’s second law and initial conditions consist in assigning initial positionand velocity of the mass in movement. The second has also an interest, it might correspond to the problemof finding a trajectory of a mass that at some initial time t0 is in some position y0 and at some future timet1 reaches a position y1. The different nature of these problems is also reflected by different results wemay prove concerning existence and uniqueness. This holds true in general for the Cauchy Problem, butmight be false for the Boundary Value Problem.

Theorem 1.4.5. The Cauchy Problem CP(t0, y0, y′0) has a unique solution for any t0 ∈ I and y0, y

′0 ∈ R.

Proof — It is easy: we have to prove that there exists a unique c1, c2 such that

y = c1w1 + c2w2 +U,

is a solution of CP(t0, y0, y′0). Just impose the two conditions: we get

c1w1(t0) + c2w2(t0) +U(t0) = y0,

c1w′1(t0) + c2w

′2(t0) +U ′(t0) = y′0,

⇐⇒

c1w1(t0) + c2w2(t0) = y0 −U(t0),

c1w′1(t0) + c2w

′2(t0) = y′0 −U ′(t0).

Now, look at this as a system 2 × 2. The coefficient matrix is the wronskian matrix which is, in our assumption,invertible. Therefore the system has a unique solution c1, c2.

Example 1.4.6. Find the solution of the Cauchy Problemy′′(t) + y(t) = et, t ∈ R,y(0) = 0,y′(0) = 1.

Sol. — The characteristic equation is λ2 + 1 = 0, that is λ = ±i. Therefore w1(t) = cos t, w2(t) = sin t is afundamental system of solutions for the homogenous equation. The wronskian is

W(t) = det

cos t sin t

− sin t cos t

= (cos t)2 + (sin t)2 = 1.

16

Therefore a particular solution, by the Lagrange formula, is

U(t) = −(∫

sin t1

et dt)

cos t +(∫

cos t1

et dt)

sin t = −(∫

et sin t dt)

cos t +(∫

et cos t dt)

sin t

= −et

2(sin t − cos t) cos t +

et

2(cos t + sin t) sin t =

et

2.

Hence the general integral is

ϕ(t) = c1 cos t + c2 sin t +et

2.

Now, imposing the initial conditions we get the systemc1 +

12 = 0,

c2 +12 = 1,

⇐⇒ c1 = −12, c2 =

12, =⇒ ϕ(t) =

12

(sin t − cos t + et

).

We close this Section by illustrating few examples of applicative use of second order linear equations.

Example 1.4.7 (Damped Oscillations). Consider the equation of motion of a mass m subjected to anelastic force (of elastic constant κ) on a viscous media (viscosity ν):

mx ′′(t) = −κx(t) − νx ′(t), ⇐⇒ mx ′′(t) + νx ′(t) + κx(t) = 0.

Describe the behavior of the solutions.Sol. — The equation is just a linear second order equation with constant coefficients. Its characteristic equation is

mλ2 + νλ + κ = 0.

Because ∆ = ν2 − 4mκ we have that if ∆ > 0, that is if ν2 > 4mκ, ν >√

4mκ, the fundamental solutions are ofexponential type, so we haven’t oscillations. Also as ∆ = 0 we have the same. To have oscillations we need ∆ < 0,that is ν <

√4mκ. In this case

λ± = −ν

2m± i

√−∆

2m,

and the fundamental system of solutions is

w+(t) = e−ν

2m t cos

(√−∆

2mt

), w−(t) = e−

ν2m t sin

(√−∆

2mt

).

In this case the oscillations are attenuated by the exponential e−ν

2m t that goes to 0 as t −→ +∞. The heavier themass is, the stronger the attenuation is.

Example 1.4.8 (Resonance). Describe long time behavior of the solutions of

(1.4.7) y′′(t) = −k2y(t) + sin(kt).

This equation is often used as model for the so–called phenomenon of resonance. For instance it wasused in the case of Tacoma bridge replacing the equation for the angle θ(t) (non linear) with an its linearapproximation. The interesting aspect of this equation is that presents unbounded solutions.

17

Sol. — The characteristic equation is λ2 = −k2, that is λ = ±ik, therefore the fundamental system of solutionsfor the homogeneous equations is w1(t) = cos(kt), w2(t) = sin(kt). The wronskian is W(t) ≡ k and a particularsolution is

U(t) = −(∫

sin(kt)k

sin(kt) dt)

cos(kt) +(∫

cos(kt)k

sin(kt) dt)

sin(kt).

Now∫sin(kt)2 dt =

∫sin(kt) sin(kt) dt = −

1k

∫sin(kt) (cos(kt))′ = −

1k

[sin(kt) cos(kt) − k

∫cos(kt)2 dt

]= −

12k

sin(2kt) +∫(1 − sin(kt)2) dt = −

12k

sin(2kt) + t −∫

sin(kt)2 dt

and by this we have ∫sin(kt)2 dt =

t2−

sin(2kt)4k

.

Moreover ∫cos(kt)

ksin(kt) dt =

12k

upupupupup sin(2kt) dt = −cos(2kt)

4k2 .

In conclusion

U(t) =(sin(2kt)

4k2 −t

2k

)cos(kt) −

cos(2kt)4k2 sin(kt).

By this the conclusion is evident.

Figure 1. Plot of U.

1.5. ExercisesExercise 1.5.1. Find the general integral of the following equations:

1. y′ + (cos t)y = 12 sin(2t), t ∈ R. 2. y′ − t

1−t2 y = t, t ∈] − 1,1[. 3. y′ + 2ty = 2t3, t ∈ R.

4. y′ − 1t y +

log tt = 0, t ∈]0,+∞[. 5. y′ + (tan t)y = t3, t ∈

]− π2 ,

π2[. 6. y′ + 2ty = te−t

2, t ∈ R.

7. y′ + y = sin t, t ∈ R. 8. y′ + (cos t)y = (cos t)2, t ∈ R. 9. y′ = 2tt2+1 y + 2t(t2 + 1), t ∈ R.

Exercise 1.5.2. Solve the Cauchy Problem y′(t) + 3t2

t3+5 y(t) =3√t,

y(0) = 1.

18

Exercise 1.5.3. Consider the equation

y′ − (tan t)y =1

sin t, t ∈

]0,π

2

[.

i) Find the general integral. ii) Is it true that for every solution limt→0+ y(t) = −∞ ? iii) Are there solutions suchthat ∃ limt→ π

2 −y(t) ∈ R. In this case, what is the value of the limit?


y′ + (sin t)y = sin t, t ∈ R.

i) Find its general integral. ii) Are there solutions y such that ∃ limt→+∞ y(t) ∈ R. iii) Find the solution of theCauchy Problem y

(π2)= 1.


y′(t) = −1ty(t) + arctan t .

Find its general integral on ] − ∞,0[ and on ]0,+∞[. Does it exists a y : R −→ R solution on both ] − ∞,0[ and]0,+∞[. In this case, what is y(0)?

Exercise 1.5.6. Solve the Cauchy problems

1.

y′ =

y2 − y − 23

arcsin t,

y(0) = 3.

2.

y′ =

y√

2y − 1cosh t

.

y(0) = 1.

3.

y′ =

cos2(2y)t(2 − log2 t)

y(1) = π2 .

4.

y′ =

(et + 1)y√

1 − y

et + 2,

y(0) = 1/2.

Exercise 1.5.7. Find, in function of the initial condition y(0) = y0 the solution of the Cauchy problemy′ = 4y(1 − y),

y(0) = y0.

Plot quickly a qualitative graph of the various solutions.

Exercise 1.5.8. Consider the Cauchy problem y′ = y(1 − y2),

y(0) = 1/2.

Determine the implicit form for the solution. Is it true that the solution is defined for all times t ∈ R?

Exercise 1.5.9. For each of the following equations find a fundamental system of solutions and write the generalintegral.

1. y′′ − 3y′ + 2y = 0. 2. y′′ − 2y′ + 2y = 0. 3. y′′ − 4y + 3y = 0. 4. y′′ + y′ = 0. 5. y′′ − y′ + y = 0.

Exercise 1.5.10. Find the general integral of the following equations:

1. y′′(t) + y′(t) − 6y(t) = 2e−t . 2. y′′ − y′ + y = et . 3. y′′ + 4y′ + 2y = t2. 4. y′′ + 2y′ = et .

5. y′′ − y = cos t . 6. y′′ + y = 1cos t . 7. y′′ + 2y′ + 2y = 2t + 3 + e−t . 8. y′′ − 2y′ + 2y = et cos t .

19

Exercise 1.5.11. For each of the following equations find the general integral and the solution of the CauchyProblem with initial conditions y(0) = y′(0) = 0.

1. y′′ − y = t . 2. y′′ + 4y = et . 3. y′′ + y = t.

4. y′′ + y′ − 6y = −4et . 5. y′′ − 8y′ + 17y = 2t + 1. 6. y′′ + y = 1cos t .

Exercise 1.5.12. Consider the following differential equationy′′(t) − y′(t) = tet, t ∈ R.

i) Find its general integral. ii) Are there solutions such that limt→+∞ y(t) ∈ R? iii) Find the solution of the CauchyProblem y(0) = 1, y′(0) = 0.

Exercise 1.5.13. Find the general integral of the equation

y′′(t) − 5y′(t) − 6y(t) = 16e−2t, t ∈ R.

Hence, say if there exists a solution such that y(0) = 0 and limt→+∞ y(t) = 0.

Exercise 1.5.14. Consider the equationy′′(t) + y′(t) = t + cos t, t ∈ R.

Find its general integral. Say it there are solutions y such that ∃ limt→+∞ y(t) ∈ R. Say it there are solutions y suchthat y(0) = 0 and limt→−∞ y(t) = +∞.


y′′(t) + y(t) =1

cos t, t ∈

]−π

2,π

2

[.

Find its general integral. Is it true that any solution of the equation is such that limt→ π2 −

y(t) = +∞? Say if thereare solutions such that y(t) ∼ Ct2 per t → 0 (for some C , 0).

Exercise 1.5.16. Find the general integral of the equation

y′′(t) + 4y′(t) + 4 =e−2t

t2 , t ∈]0,+∞[.

Are there solutions of the equation such that ∃ limt→0+ y(t)?

Exercise 1.5.17. A radioactive material decay of 20% in 10 days. Find his halving time.

Exercise 1.5.18. In an hospital a radioactive substance is accumulated into a vessel at rate of 2m3 each month.The radioactivity has a decay rate estimated to be proportional to the quantity present in the vessel according aconstant of proportionality k = −1. Knowing that initially the vessel is empty find the total amount of radioactivesubstance contained when the vessel is full.

Exercise 1.5.19. The water in a pool with squared base of side 10m and depth 2m evaporates at rate of 5 literseach hour. The bottom of the pool is porous with a certain speed ph expressed in liter/minute, where p is a constantand h is the level of the water into the pool. Once the pool is full of water, it takes 24 hours in order to have thepool completely empty. The problem is: determine p.

Exercise 1.5.20. A vessel has capacity 10 liters. A valve is opened on the bottom of the vessel with flow per hourproportional with constant 3 to the 3

√ of the total volume present inside the vessel. If initially the vessel is full,compute how long it takes the vessel to be completely empty. Suppose now that a constant flux F of fluid is infusedinto the vessel. Are there values of F such that the fluid into the vessel reach, at long times, an equilibrium?

20

Exercise 1.5.21. The queue formed after a car accident on an highway reduces at some rate inversely proportionalto the square root of the length of the queue with some constant c of proportionality. Knowing that to reduce to thehalf a queue of 1km it takes 10min, how long it takes to reduce to the half a queue of 2km? Do a queue reduces to0 in a finite time?

Exercise 1.5.22. In a fish breeding the population of fishes is assumed to follows a logistic evolutiony′(t) = 0.1y(t) − by(t)2,

where b is to be determined. You know that initially there’re 500kg of fishes and after one year there’re 1.250kg offishes. Determine b.

Exercise 1.5.23 (?). A particle of mass m fall down under action of gravity and air friction in such a way that theequation of motion is

ma(t) = −mg − mkv(t).Find an equation for v as function of the quote x and find v = v(x). This is the case of a light particle. If we havean heavy particle the friction changes as −mkv(t)2. What can you say in this case?

Exercise 1.5.24 (??). A swimmer want to cross a river of section `. He starting point and the arriving point arealigned orthogonally to the direction of the river. The water into the river flow at constant speed v. The swimmerwant to follow a trajectory always directed to his destination with constant speed V < v. Does he will reach theother side of the river? Describe the trajectory of the swimmer through a suitable differential equation and findunder which conditions the swimmer we be succeful.

Exercise 1.5.25 (?). A ship of mass m moves from rest under a constant propelling force m f and against aresistance mkv2. Determine the speed v = v(a) as function of the covered distance a. Suppose that, fixed a, theengines are reversed. What is the distance necessary to stop the ship?

Exercise 1.5.26. A particle of unit mass moves along the x−axis under the attraction of a forse of magnitude 4xtowards the point x = 0 and a resistance equal in magnitude to twice the velocity. The particle is released at rest atx = a. Determine all the positions at instantaneous rest.

Exercise 1.5.27. A mass m is attached to two springs along the vertical with same elastic constant k. Initially themass is at rest positions for the springs. Determine the motion of the mass taking account of gravity.

CHAPTER 2

Topology in Rd

In a wide part of this course we will study Analysis in a multidimensional context,

Rd := {(x1, . . . , xd) : xi ∈ R, i = 1, . . . , d} .

The basic tool ofMathematical Analysis is the concept of limitwith its applications: continuous functions,differential calculus, integral calculus andmore. The definition of limit passes through a suitable definitionof distance between points. In Rd there’s a natural notion of distance, namely the Euclidean distance

dist(x, y) =√(x1 − y1)2 + . . . + (xd − yd)2, x = (x1, . . . , xd), y = (y1, . . . , yd).

The euclidean distance is actually a function of the difference between the coordinates of the two pointsx, y or, in other words, is translation invariant. The distance d(x,0) (where 0 = (0, . . . ,0) is the origin)can be interpreted as the length of a vector and plays the same role of the modulus for numbers. This isthe fundamental concept of norm by wich we will begin.

2.1. Euclidean norm

We start recalling that Rd is a vector space on R with the operations of sum and product

(x1, . . . , xd) + (y1, . . . , yd) := (x1 + y1, . . . , xd + yd), λ(x1, . . . , xd) := (λx1, . . . , λxd).

Definition 2.1.1. Given x = (x1, . . . , xd) ∈ R we call norm of x the quantity

‖x‖ :=

√√√d∑i=1

x2i .

The norm plays the same role of the modulus in R. Precisely

Proposition 2.1.2. The norm fulfills the following properties:i) positivity: ‖x‖ > 0, ∀x ∈ Rd, and ‖x‖ = 0 iff x = 0d := (0, . . . ,0).ii) homogeneity: ‖λx‖ = |λ |‖x‖, ∀λ ∈ R, ∀x ∈ Rd.iii) triangular inequality: ‖x + y‖ 6 ‖x‖ + ‖y‖, ∀ x, y ∈ Rd.

Proof — ‖x‖ > 0 for eny x ∈ Rd is evident. Let’s check the vanishing:

‖x‖ = 0, ⇐⇒d∑i=1

x2i = 0, ⇐⇒ x2

i = 0, ∀i, ⇐⇒ xi = 0, ∀i.

21

22

The homogeneity is very easy:

‖λx‖ =√∑

i

(λxi)2 =√λ2

∑i

x2i = |λ |

√∑i

x2i = |λ |‖x‖.

Finally the triangular inequality: for convenience let’s square everything and notice that

‖x + y‖2 =∑i

(xi + yi)2 =

∑i

(x2i + y2

i + 2xiyi) = ‖x‖2 + ‖y‖2 + 2∑i

xiyi .

Lemma 2.1.3 (Cauchy–Schwarz inequality).

(2.1.1)∑i

xiyi 6

(∑i

x2i

)1/2 (∑i

y2i

)1/2

.

Proof — (Lemma) Excluding the trivial cases when ‖x‖ = 0 or ‖y‖ = 0 we may assume ‖x‖, ‖y‖ , 0 and prove∑i

xi‖x‖

yi

‖y‖6 1.

This simple trick make easy to conclude: recall the elementary inequality

ab 612(a2 + b2),

(⇐⇒ 2ab 6 a2 + b2, ⇐⇒ (a − b)2 > 0

).

Then using a = xi‖x ‖ and b = yi

‖y ‖ we have∑i

xi‖x‖

yi

‖y‖6

12

∑i

(x2i

‖x‖2+

y2i

‖y‖2

)=

12

(‖x‖2

‖x‖2+‖y‖2

‖y‖2

)= 1.

By Cauchy–Schwarz,‖x + y‖2 6 ‖x‖2 + ‖y‖2 + 2‖x‖‖y‖ = (‖x‖ + ‖y‖)2 , ⇐⇒ ‖x + y‖ 6 ‖x‖ + ‖y‖.

Through the norm we define the concept of limit for a sequence and, later, the limit of a function:

Definition 2.1.4. Let (xn) ⊂ Rd be a sequence of vectors. We say thatxn −→ ξ ∈ Rd, ⇐⇒ ‖xn − ξ‖ −→ 0 (in R).

A little bit of care is needed for xn −→ ∞ because, differently by R, there’s not a +∞ and a −∞:

Definition 2.1.5. Let (xn) ⊂ Rd be a sequence of vectors. We say thatxn −→ ∞d, ⇐⇒ ‖xn‖ −→ +∞.

A neighborhood, that is a set of points closed to a given point x:

Definition 2.1.6. Let x ∈ Rd and r > 0. We callclosed ball centered in x of radius r the set B(x,r] := {y ∈ Rd : ‖y − x‖ 6 r},

open ball centered in x of radius r the set B(x,r[:= {y ∈ Rd : ‖y − x‖ < r}.

Every set Ux containing a ball centered at x ∈ Rd its called neightborhood of x. We call alsoneighborhood of∞d any set U∞ containing the exterior of a ball, that is U∞d

⊃ B(0,r]c.

23

Remark 2.1.7. As d = 1 we have B(x,r] = [x − r, x + r]; as d = 2, B(x,r] is the disk centered in x with radius r .We used the square bracket in the notation B(x,r] to recall that B(x,r] contains all the points distant exactly r fromthe center (that is the "skin" of the ball).

Definition 2.1.8 (open set). A set S ⊂ Rd is said• open, if ∀ x ∈ S ∃ r(x) > 0 : B(x,r(x)[⊂ S.• closed, if its complementary Sc = Rd\S is open.

By definition ∅ is assumed to be open.

Remark 2.1.9. A common error consists in thinking that every set S is open or closed. This a wrongidea probably due to the meaning of "closed" and "open" in the common language. For instance:

• in R, [a, b[ is neither open or closed;• ∅ is open by definition; according to the definition it is also closed: indeed ∅c = Rd which isclearly open. Same for Rd

There are no other simultaneously closed and open subsets in Rd, but this is not easy to prove.

Proposition 2.1.10. If A,B ⊂ Rd are open (closed) sets then A ∪ B, A ∩ B are open (closed).Proof — Exercise.

An important characterization of closed sets is the following:

Theorem 2.1.11 (Cantor). S is closed iff it contains all the finite limits of all its sequences that is(2.1.2) S closed ⇐⇒ ∀(xn) ⊂ S, : xn −→ ξ ∈ Rd, then ξ ∈ S.

Proof — =⇒ Assume that S is closed and let’s prove that if (xn) ⊂ S with xn −→ ξ ∈ Rd then ξ ∈ S. Assume thatthis is false: then ξ ∈ Sc . But Sc is open (being S closed) and because ξ ∈ Sc ,

∃ B(ξ,r[⊂ Sc .

But xn −→ ξ, that is ‖xn − ξ‖ −→ 0 hence ‖xn − ξ‖ < r definitively: this means that xn ∈ B(ξ,r[⊂ Sc definitively,and this is a contradiction being (xn) ⊂ S.

⇐= Assume the the property (2.1.2) is true and let’s prove that S is closed, that is Sc is open. Take ξ ∈ Sc andassume that, by contradiction,

� B(ξ,r[⊂ Sc .

Then∀r > 0, B(ξ,r[1 Sc, ⇐⇒ ∀r > 0, ∃x ∈ B(ξ,r[ : x ∈ S.

Take r = 1n and call xn ∈ B(ξ, 1

n [ such that xn ∈ S. The sequence (xn) ⊂ S and because ‖xn − ξ‖ < 1n −→ 0 we

deduce xn −→ ξ. But then, by (2.1.2), ξ ∈ S, and this is a contradiction.

Let’s introduce two useful concepts:

Definition 2.1.12 (interior and boundary). Let S ⊂ Rd. We call• Int(S) (interior of S) the set of points of x ∈ S such that there exists Ux (neighborhood of x)such that Ux ⊂ S;

24

• ∂S (boundary of S) the set of points x ∈ Rd such that every neighborhood Ux of x containspoint of S and of Sc, that is Ux ∩ S , ∅, Ux ∩ Sc , ∅.

In particular, S is open if and only if S = Int(S).

2.2. Limit

In this section we want to define the notion of limitlimx→x0

f (x) = `,

for a function f : D ⊂ Rd −→ Rm. As in one variable Calculus, to set this Defnition we need the conceptaccumulation point:

Definition 2.2.1. Let S ⊂ Rd. We say that• ξ ∈ Rd is accumulation point for S if ∃(xn) ⊂ S\{ξ} such that xn −→ ξ;• ∞d is accumulation point for S if ∃(xn) ⊂ S such that xn −→ ∞d.

The set of all accumulation points of S will be denoted by Acc(D).

By this and importing the same idea introduced for limits of one real variable functions we have the:

Definition 2.2.2. Let f : D ⊂ Rd −→ Rm and x0 ∈ Acc(D). We say that(2.2.1) lim

x→x0f (x) = ` ∈ Rm ∪ {∞m}, ⇐⇒ f (xn) −→ `, ∀(xn) ⊂ D\{x0}, xn −→ x0.

This Definition has the advantage to cover all the possibilities: limit at a finite point (when x0 ∈ Rd), at

infinite (when x0 = ∞d) as well as finite limit (when ` ∈ Rm) or infinite limit (` = ∞m). Despite this, theDefinition is not helpful to compute practically a limit. Let’s see some useful techniques.

2.2.1. Sections. Let f : D ⊂ Rd −→ Rm be a function such that limx→x0 f (x) = `. Wemay imaginethat taking a "road" into D going to x0, f will drive us just to `. With "road" we mean a line in the space.

Definition 2.2.3 (curve). A function γ : [a, b] −→ Rd is called curve in D if γ(t) ∈ D for everyt ∈ [a, b] (notation γ ⊂ D). We call support of γ the set Supp(γ) := γ([a, b]). The curve γ is said to becontinuous if γ ∈ C ([a, b]).

Proposition 2.2.4. Let f : D ⊂ Rd −→ Rm, x0 ∈ Acc(D) be such that limx→x0 f (x) = `. Let γ ⊂ D bea curve such that limt→t0 γ(t) = x0 and γ(t) , x0 forall t. Then

limt→t0

f (γ(t)) = `.

Proof — It’s just an application of the definitions. Take tn −→ t0: then, becauselimt→t0

γ(t) = x0, =⇒ γ(tn) −→ x0.

By iii) we know that xn := γ(tn) , x0 and we come to see that xn −→ x0. Therefore, by the Definition (2.2.1)f (xn) = f (γ(tn)) −→ `. So we proved that

∀tn −→ t0, =⇒ f (γ(tn)) −→ `,

and this is nothing but the conclusion.

25

Corollary 2.2.5. Let f : D ⊂ Rd −→ Rm, x0 ∈ Acc(D). If there exists γ1, γ2 curves in D fulfilling ii)and iii) of the Proposition 2.2.4 and such that

limt→t0

f (γ1(t)) , limt→t0

f (γ2(t))

then the limx→x0 f (x) doesn’t exists.

Example 2.2.6. Show thatlim

(x,y)→02

xyx2 + y2

doesn’t exists.Sol. — Let

f (x, y) =xy

x2 + y2 , (x, y) ∈ D = R2\{(0,0)}.

Let’s check what happens along the two sections along the axes. These are

f (t,0) = 0, f (0, t) = 0.

Here γ1(t) = (t,0) −→ (0,0) as t −→ 0 =: t0 and clearly γ1(t) , (0,0) for all t , t0. Hence

f (γ1(t)) = f (t,0) = 0 −→ 0, as t −→ 0.

Similarly f (0, t) −→ 0. Is this enough to conclude that the limit exists? NO! Because we checked just two of theinfinitely many sections. Let consider a new section, that is a point moving along a straight line y = mx. The curvedescribing this is simply

γ(t) := (t,mt), m ∈ R.Notice that the corresponding section of f is

f (γ(t)) = f (t,mt) =mt2

t2 + m2t2 =m

1 + m2 −→m

1 + m2 , as t −→ 0.

We conclude that the behavior of f along the axes is different to that one along straight lines through the originwith angular coefficient m , 0. The limit doesn’t exists.

Example 2.2.7. Show that

lim(x,y)→(0,0)

xy2

x2 + y4

doesn’t exists.Sol. — Let

f (x, y) =xy2

x2 + y4 , (x, y) ∈ D = R2\{(0,0)}.

The sections along the axes are f (t,0) ≡ 0 and f (0, t) ≡ 0. Notice that this says, in particular, that if the limitexists, it must be equal to 0. Now if we take a section along the line y = mx,

f (t,mt) =m2t3

t2 + m2t4 =m2t

1 + m2t2 −→ 0, as t −→ 0.

So apparently again no contradictions! But if we consider the line x = ay2 we have

f (at2, t) =at2t2

a2t4 + t4 =a

a2 + 1−→

aa2 + 1

, as t −→ 0.

26

This is different from 0 if a , 0: so we have found a family of curves on which the limit of f exists but is differenton any family: we deduce that the limit doesn’t exists.

Example 2.2.8. Show thatlim

(x,y)→∞2

(x2 + y2 − 4xy

)doesn’t exists.Sol. — Let f (x, y) := x2 + y2 − 4xy. Sections along the axes are f (t,0) = t2, f (0, t) = t2. Clearly the points(t,0), (0, t) go to ∞2 iff t −→ ±∞. In any case f (t,0), f (0, t) −→ +∞. So the candidate to be the eventual limit is+∞. However, along the line y = x,

f (t, t) = t2 + t2 − 4t2 = −2t2,

and because (t, t) −→ ∞2 iff t −→ ±∞we have immediately that f (t, t) −→ −∞. We conclude that the limit doesn’texists.

We have seen then that sections may be used to• guess the possible limit (because if the limit exists then along any section the limit exists and itis the same);• exclude existence of the limit (if there’re two different sections alongwhich the limits are differentthe global limit cannot exists).

Of course to guess what the "right" sections are is not an easy business.

2.2.2. Methods of calculus for scalar functions. Sections are useful to find a candidate or to excludeexistence of the limit, but are useless to prove that a function has a limit. In the following Example wewill introduce an interesting method to answer to this problem.

Example 2.2.9. Compute

lim(x,y)→02

xy2

x2 + y2 .

Sol. — We have to begin with to guess a candidate. We remember that if the limit exists must coincide with thelimit along any section. Now f (x,0) = 0 −→ 0, so if the limit exists must be 0. This is confirmed, by the way, bythe y−axis section f (0, y) = 0 and by sections along y = mx, because

f (x,mx) =xm2x2

x2 + m2x2 = xm2

1 + m2 −→ 0, x −→ 0.

Ok, if the limit exists it must be 0. How can we check that this is actually the case? Notice that by using polarcoordinates

x = ρ cos θ,

y = ρ sin θ,we have

f (ρ cos θ, ρ sin θ) =ρ3(cos θ)(sin θ)2

ρ2 = ρ(cos θ)(sin θ)2,

so| f (ρ cos θ, ρ sin θ)| 6 |ρ(cos θ)(sin θ)2 | 6 ρ.

27

Returning to euclidean coordinates this last says that| f (x, y)| 6 ‖(x, y)‖.

It seems now evident that if (x, y) −→ 02, that is ‖(x, y)‖ −→ 0+ then, by some argument similar to the two-policemen thm, we should have also | f (x, y)| −→ 0+, that is f (x, y) −→ 0.

What is the argument invoked at the end of the previous example? We found a numerical functiong = g(ρ) : [0,+∞[−→ R such that g(ρ) −→ 0 as ρ −→ 0 and

| f (x, y)| 6 g(‖(x, y)‖).

It is clear that, as ‖(x, y)‖ −→ 0 (that is (x, y) −→ 02) by the two policemen Lemma we have easily thatf (x, y) −→ 0. We can extend this to a more general setting:

Proposition 2.2.10. Let f : D ⊂ Rd −→ R, ξ ∈ Acc(D). Suppose that there exist g such thati) | f (x) − ` | 6 g(‖x − ξ‖) in some Uξ\{ξ};ii) limρ→0+ g(ρ) = 0.

Then ∃ limx→ξ f (x) = `.Proof — Let xn −→ ξ, that is ‖xn − ξ‖ −→ 0. Then

| f (xn) − ` | 6 g(‖xn − ξ‖) −→ 0, =⇒ f (xn) −→ `.

Example 2.2.11. Compute

lim(x,y,z)→03

sin(xyz)x2 + y2 + z2 .

Sol. — Let f (x, y, z) := sin(xyz)x2+y2+z2 defined on its natural domain D = R3\{03}. The sections on the axes

f (x,0,0) = f (0, y,0) = f (0,0, z) vanish, so the eventual candidate to be the limit is 0. Using spherical coordinatesx = ρ cos θ sin ϕ,y = ρ sin θ sin ϕ,z = ρ cos ϕ,

we have

f (ρ cos θ sin ϕ, ρ sin θ sin ϕ, ρ cos ϕ) =sin

(ρ3(cos θ)(sin θ)(sin ϕ)2(cos ϕ)

)ρ2 .

Recalling that sin(ξ) = ξ + o(ξ) we have

f (ρ cos θ sin ϕ, ρ sin θ sin ϕ, ρ cos ϕ) =ρ3(cos θ)(sin θ)(sin ϕ)2(cos ϕ)

ρ2 +o(ρ3(cos θ)(sin θ)(sin ϕ)2(cos ϕ))

ρ2

= ρ(cos θ)(sin θ)(sin ϕ)2(cos ϕ) + o(ρ(cos θ)(sin θ)(sin ϕ)2(cos ϕ)

).

Clearly,|ρ(cos θ)(sin θ)(sin ϕ)2(cos ϕ)| 6 ρ −→ 0, as ρ −→ 0+,

hence o(. . .) −→ 0. Therefore the limit exists and is 0.

We have a similar strategy in the case ` = +∞ (or −∞):

28

Proposition 2.2.12. Let f : D ⊂ Rd −→ R, ξ ∈ Acc(D). Suppose that there exist g such thati) | f (x)| > g(‖x − ξ‖) in some Uξ\{ξ};ii) limρ→0+ g(ρ) = +∞.

Then ∃ limx→ξ f (x) = +∞.Proof — Exercise.

A final important case is when x −→ ∞d. We just quote the following

Proposition 2.2.13. Let f : D ⊂ Rd −→ R,∞d ∈ Acc(D). Suppose that there exist g such thati) f (x) > g(‖x‖) in some U∞d

;ii) limρ→+∞ g(ρ) = +∞.

Then ∃ limx→∞df (x) = +∞.

Proof — Exercise.

Example 2.2.14. Computelim

(x,y)→∞2

(x4 + y4 − xy

).

Sol. — Looking at the sections along the axes we have f (x,0) = x4 −→ +∞ and f (0, y) = y4 −→ +∞. So, if thelimit exists must be +∞. This seems reasonable because x4 + y4 should dominate xy. In this case we need just a"lower" policemen g = g(ρ) such that

f (ρ cos θ, ρ sin θ) > g(ρ) −→ +∞, ρ −→ +∞.

We have

f (ρ cos θ, ρ sin θ) = ρ4(cos θ)4 + ρ4(sin θ)4 − ρ2(cos θ)(sin θ) = ρ4 [(cos θ)4 + (sin θ)4

]−

12ρ2 sin(2θ).

Now: notice that the quantity K(θ) := (cos θ)4 + (sin θ)4 is always positive and has a minimum as θ ∈ [0,2π].Indeed: we don’t need any computation because K is clearly continuous, hence K has a minimum by Weierstrass’stheorem. Moreover K(θ) = 0 iff cos θ = sin θ = 0, and this in impossible. We call C the minimum value of K:K(θ) > C > 0 for any θ ∈ [0,2π]. Recalling also that | sin(2θ)| 6 1 we have

f (x, y) > Cρ4 −12ρ2 sin(2θ) > Cρ4 −

12ρ2 =: g(ρ) −→ +∞.

By this the conclusion follows.


(x,y,z)→∞3

[(x2 + y2 + z2)2 − xyz

].

Sol. — A quick check on the sections along the axes show that they tend to +∞. Again: it seems reasonable thatthe fourth order term (x2 + y2 + z2)2 dominates on xyz. Passing to spherical coordinates

f = (ρ2)2 − ρ3(cos θ)(sin θ)(sin ϕ)2(cos ϕ) = ρ4 −14ρ3(sin(2θ))(sin(2ϕ))(sin ϕ).

Now, because|(sin(2θ))(sin(2ϕ))(sin ϕ)| 6 1,

29

we havef > ρ4 −

14ρ3 =: g(ρ) −→ +∞,

from which the conclusion follows.


(x,y,z)→∞3

[(x2 + y2)2 + z2 − xy

]Sol. — Easily the sections are all convergent to +∞ (e.g. f (x,0,0) = x4 −→ +∞ when ‖(x,0,0)‖ = |x | −→ +∞).In this case it is convenient to introduce cylindrical coordinates

x = ρ cos θ,y = ρ sin θ,z = z

because x2 + y2 = ρ2. But be careful: (x, y, z) −→ ∞3 means ‖(x, y, z)‖ =√

x2 + y2 + z2 =√ρ2 + z2 −→ +∞,

and this doesn’t mean necessarily that ρ −→ +∞. However,fcil = (ρ2)2 + z2 − ρ2 cos θ sin θ > ρ4 + z2 − ρ2, (| cos θ sin θ | 6 1).

Now: if we had f (x, y, z) > ρ2 + z2 = ‖(x, y, z)‖2 we would be done. To this aim we may hope that ρ4 − ρ2 > ρ2

and indeed this is actually true if ρ is big enough but not for every ρ. To get a lower bound true for any ρ we maynotice that

∃K : ρ4 − ρ2 > ρ2 + K, ∀ρ.Indeed: this is equivalent to say that ρ4−2ρ2 > K , that is the function ρ 7−→ ρ4−2ρ2 is bounded below. But a quickcheck shows that this function has a global minimum: so, if we call K the minimum of the function ρ 7−→ ρ4 − 2ρ2

we have the conclusion.

2.3. Continuity

One of the major application of the concept of limit is the definition of continuity:

Definition 2.3.1 (Continuous function). Let f : D ⊂ Rd → Rm, x0 ∈ D ∩ Acc(D). We say thatf is continuous in x0 iff lim

x→x0f (x) = f (x0).

If f is continuous in any point of D we say that f is continuous over D and we write f ∈ C (D).

The usual properties of continuity for one variable functions remain true. For instance: sum, differenceand products of continuous functions at some point (or in some domain) is a continuous function at thatpoint (or in that domain). The same for the ratio with the extra requirement that the denominator isdifferent from 0. It is quite easy to prove (we omit this) that

any polynomial in the (x1, . . . , xd) variable is continuous on Rd .By polynomial we mean a finite sum of monomials of type

axk11 xk2

2 · · · xkdd, k1, . . . , kd ∈ N, a ∈ R.

Quite useful is the chain rule:if f is continuous in x0, g is continuous in f (x0) then g ◦ f is continuous in x0.

30

For instance: any continuous scalar function of a polynomial is continuous where defined.

Example 2.3.2. Where is continuous the function f (x, y) := log(1 − x2 − y2)?Sol. — The function is defined on

D ={(x, y) ∈ R2 : 1 − x2 − y2 > 0

}=

{(x, y) ∈ R2 : x2 + y2 < 1

}= B(0,1[.

We may write f = log ◦p where p is the polynomial p(x, y) := 1 − x2 − y2. Therefore f ∈ C (B(0,1[).

Example 2.3.3. The euclidean norm is continuous on Rd.Sol. — Remind that

‖x‖ =√

x21 + . . . + x2

d≡√◦ p, where p(x1, . . . , xd) = x2

1 + . . . + x2d .

Now: √ is continuous where defined and p > 0. It follows ‖ · ‖ ∈ C (Rd).

It is easy to check that continuity component wise:

Proposition 2.3.4. Let f : D ⊂ Rd −→ Rm, f = ( f1, . . . , fm). Then f is continuous in x0 iff any fj iscontinuous in x0, j = 1, . . . ,m.

In particular, if A is a linear transformation, that is

x 7−→ Ax =

a11 . . . a1d...

...am1 . . . amd

©«

x1...

xd

ª®®¬ =©«

a11x1 + . . . + a1dxd...

am1x1 + . . . + amdxd

ª®®¬ ,because every component is a first order polynomial, we get by the last proposition that A is continuous:

Corollary 2.3.5. Any linear transformation T ∈ L (Rd;Rm) is continuous.

2.4. Properties of continuous functions

2.4.1. Weierstrass Theorem. We recall that any f ∈ C ([a, b];R) has minimum/maximum over[a, b]. This is the well known Wierstrass Thm. The conclusion is false if the interval [a, b] is not closedand bounded. These two properties are the key properties to extend the Thm to the case of functions ofseveral variables. We need first to give the

Definition 2.4.1. A set S is said bounded if

∃M, : ‖x‖ 6 M, ∀x ∈ S.

Theorem 2.4.2 (Weierstrass). Any continuous function f : D ⊂ Rd −→ R on a domain D closed andbounded has global minimum and global maximum on D, that is

∃ xmin, xmax ∈ D, : f (xmin) 6 f (x) 6 f (xmax), ∀x ∈ D.

Weierstrass thm points out the importance of the class of closed and bounded subsets of Rd:

Definition 2.4.3. A set S ⊂ Rd is called compact if it is closed and bounded.

31

Sowemayquickly say that continuous functions on compact sets havemin/max. Ifwe remove compactnesswe cannot assure the existence of global extreme points. There’re however cases when the domain D isstill closed but unbounded (as for instance when D = Rd) in which something can be said. We notice that

S unbounded ⇐⇒ ∀n ∃xn ∈ S : ‖xn‖ > n, ⇐⇒ ∃(xn) ⊂ S, xn −→ ∞d .

In particular, S is unbounded iff∞d ∈ Acc(S).

Corollary 2.4.4. Let f : D ⊂ Rd −→ R be continuous on D, closed and unbounded, such thatlim

x→∞d

f (x) = +∞ (−∞).

Then f has a global minimum (maximum).Proof — Fix a point x0 ∈ D: by hypotheses, there exists R such that

f (x) > f (x0) + 1, ∀x ∈ D : ‖x‖ > R.

Indeed: if such R wouldn’t exists, for any R = n ∈ N then there should be a point xn ∈ D such that f (xn) 6 f (x0)+1and ‖xn‖ > R = n. But then xn −→ ∞d hence by assumption f (xn) −→ +∞ which is impossible beingf (xn) 6 f (x0) + 1 (that is bounded).

Now, with such R we can notice that if the minimum exists it must belongs to D ∩ B(0,R] and also thatx0 ∈ B(0,R] (otherwise f (x0) should be grater than f (x0) + 1 which is impossible). But D ∩ B(0,R] is closed(intersection of two closed set) and bounded (because contained in B(0,R]). Therefore, by Weierstrass’s theoremapplied to f on D ∩ B(0,R], there exists xmin ∈ D ∩ B(0,R] such that

f (xmin) 6 f (x), ∀x ∈ D ∩ B(0,R].In particular, also, f (xmin) 6 f (x0) < f (x0) + 1 6 f (x) for all x ∈ D ∩ B(0,R]c . By this follows that

f (xmin) 6 f (x), ∀x ∈ D.

Remark 2.4.5. Of course, because limx→∞df (x) = +∞ the function cannot have a maximum!

Example 2.4.6. Show that the function f (x, y) := x4 + y4 − xy has global minimum on R2. What aboutglobal maximum?Sol. — Of course f ∈ C (R2) (because it is a polynomial) and R2 is closed (its complementary is empty, so openby definition) and unbounded. We have also seen (see Example 2.2.14) that

lim(x,y)→∞2

f (x, y) = +∞.

Therefore, by the Corollary of Weierstrass’s thm we have that there exists a global minimum for f on R2. On theother side, because f is upper unbounded (by the limit at∞2) the global maximum doesn’t exists.

2.4.2. Domains defined by continuous functions. Weiertrass’ Thm shows how important is to sayif a set S is closed or less. A natural way to define subsets in Rd is through equalities or inequalities:S is the intersection of the closed unit ball

{x2 + y2 + z2 6 1

}with the cylinder

{(x − 1)2 + y2 6 1

4}with

axis parallel to z−axis passing at the point (1/2,0,0) with radius 12 . A quite general setting is to define a

set as intersection of sets of type{ f 6 0}, or { f = 0}, or { f < 0}.

An important question is to know if these sets are open or closed.

32

Figure 1. The set S :={(x, y, z) ∈ R3 : x2 + y2 + z2 6 1,

(x − 1

2

)2+ y2 6 1

4

}

Theorem 2.4.7. Let f : Rd −→ R be continuous on Rd. Then

i) { f < 0} is open;ii) { f 6 0} and { f = 0} are closed.

Proof — i) Let’s show that { f < 0}c = { f > 0} is closed. To this aim let’s use the characterization (2.1.2): let(xn) ⊂ { f > 0} such that xn −→ ξ ∈ Rd . The goal is to show that ξ ∈ { f > 0}. We know that

(xn) ⊂ { f > 0}, =⇒ f (xn) > 0.

But f is continuous at ξ hence f (xn) −→ f (ξ), and because f (xn) > 0 for any n, by permanence of sign it followsthat f (ξ) > 0, that is ξ ∈ { f > 0}, and this concludes the proof.ii) The proof is similar to that one of the previous point.

Corollary 2.4.8. If f1, . . . , fk ∈ C (Rd;R) then { f1 > 0, . . . , fk > 0} and { f1 = 0, . . . , fk = 0} areclosed, { f1 > 0, . . . , fk > 0} is open.Proof — Just each of the sets is intersection of sets like { fj > 0}, { fj = 0} and { fj > 0}.

2.5. ExercisesExercise 2.5.1. Discuss the following questions:

i) B(ξ, ρ] is closed.ii) S ⊂ Rd is closed iff ∂S ⊂ S.iii) S is open iff S = Int(S).iv) Int(S) is open for every set S ⊂ Rd .v) A set S is open iff ∂S ⊂ Sc .vi) Are there cases of sets S such that Int(S) = ∅? And sets such that ∂S = ∅?vii) Prove the proposition 2.1.10.

33

Exercise 2.5.2. Looking to suitable sections, prove that the following limits don’t exist:

1. lim(x,y)→02

x2 − y2

x2 + y2 . 2. lim(x,y)→02

x2 + y3

x2 + y2 . 3. lim(x,y)→02

xy|x | + |y |

.

4. lim(x,y)→02

y2 − xyx2 + y2 . 5. lim

(x,y,z)→03

x + y2 + z3√x2 + y2 + z2

. 6. lim(x,y)→02

xy +√y2 + 1 − 1

x2 + y2 .

7. lim(x,y,z)→03

xyzx4 + y2 + z2 .

Exercise 2.5.3. Compute the following limits:

1. lim(x,y)→02

xy√x2 + y2

. 2. lim(x,y)→02

x2y3

(x2 + y2)2. 3. lim

(x,y)→02

x3 − y3

x2 + y2 . 4. lim(x,y)→02

x√|y |

3√

x4 + y4. 5. lim

(x,y)→02

xy|x | + |y |

.

Exercise 2.5.4. An open ball is an open set.

Exercise 2.5.5. For each of the following limit, say if it exists (and in the case compute it) or less:

1. lim(x,y)→02

e4y3− cos(x2 + y2)

x2 + y2 . 2. lim(x,y,z)→03

xyzx2 + y2 + z2 . 3. lim

(x,y,z)→03

(x2 + yz)2√(x2 + y2)2 + z4

.

4. lim(x,y)→02

log(1 + 2x3)

sinh(x2 + y2). 5. lim

(x,y)→(0,1)

x3 sinh(y − 1)x2 + y2 − 2y + 1

. 6. lim(x,y)→(1,1)

(x − 1)2(y − 1)7((x − 1)2 + (y − 1)2

)5/2 .

Exercise 2.5.6. For each of the following limit, say if it exists (and in the case compute it) or less:

1. lim(x,y)→∞2

(x3 + xy2 − y2

). 2. lim

(x,y)→∞2(x4 − y4 + y2 − x2).

3. lim(x,y)→∞2

(x2y2 + x2 + y2 − xy

). 4. lim

(x,y,z)→∞3

(x4 + y4 + z4 − xyz

).

5. lim(x,y,z)→∞3

(x2 + y2 + z4 − xz

). 6. lim

(x,y,z)→∞3

(√x2 + y2 + z2 − z

).

7. lim(x,y,z)→∞3

(√(x2 + y2)2 + z4 − xyz

).

CHAPTER 3

Differential Calculus

In this Chapter we extend the Differential Calculus to the general setting of vector valued functionsof several variables,

f : D ⊂ Rd −→ Rm,

where D is some domain in Rd. The case f : D ⊂ Rd −→ R is very important in view of the optimizationproblems, that is to find min/max of f on the set D. This problem is one of the main reasons to introducethe Differential Calculus because the derivative should give informations useful to search extreme points.

The extension to the multi variable setting is not at all straightforward. Just beginning with thedefinition, in our setting we can’t write

f ′(x) = limh→0

f (x + h) − f (x)h

The problem is the domain: if the dimension d > 1, x, h ∈ Rd and we do not have an operation of"division" between vectors. As we will see, there’re several possible definitions of derivative, but justprecisely one is the more appropriate.

3.1. Directional derivative

In all the section we will assume that

f : D ⊂ Rd −→ Rm, x ∈ Int(D), d > 1

Fix a vector v , 0d. The set {x + tv : t ∈ R} is the straight line passing by x and with direction v. It isclear that being x ∈ Int(D) at least for t small x + tv ∈ D (see the figure). Precisely: if B(x,r] ⊂ D thenx + tv ∈ B(x,r] iff ‖tv‖ 6 r , iff |t | 6 r

‖v ‖ . Now, in a natural way we define

x0

D

x0+tv

Definition 3.1.1. Let f : D ⊂ Rd −→ Rm, x ∈ Int(D). We call directional derivative of f in the pointx along v , 0d the limit (if it exists finite)

Dv f (x) := limt→0

f (x + tv) − f (x)t

.

35

36

The directional derivative works as an ordinary derivative but, unfortunately, is not a good concept forderivative: it may happens that all the Dv f exists but f is not even continuous!

Example 3.1.2. Let

f (x, y) :=

x2y

x4 + y2 , (x, y) , 02,

0, (x, y) = 02.

Then f has all the directional derivatives in the point 02 but it is not continuous in 02.Sol. — Let’s start by the continuity. Looking at the sections along axes we have f (x,0) = f (0, y) ≡ 0 −→ 0. Butalong the section y = x2 we have

f (x, x2) =x2x2

x4 + x4 =12−→

12, f (0,0) = 0.

Therefore� lim(x,y)→02 f (x, y) and consequently the function cannot be continuous! Let’s prove now that∃Dv f (0,0)for any v. Let v = (a, b) , 02. We have

Dv f (0,0) = limt→0

f ((0,0) + t(a, b)) − f (0,0)t

= limt→0

f (ta, tb)t

= limt→0

t3a2bt2(t2a4+b2)

t= lim

t→0

a2bt2a4 + b2 ,

that is

Dv f (0,0) =

0, if b = 0 (and of course a , 0),

a2

b2 , if b , 0.

Directional derivatives are just variations on one variable derivatives. A particularly important case isthe following:

Definition 3.1.3 (partial derivative). Let f : D ⊂ Rd −→ Rm, x ∈ Int(D) and let e1, . . . , ed be thecanonical base of Rd, that is ej = (0, . . . ,0,1,0, . . . ,0) with 1 in the j−th place. If it exists, we definepartial derivative of f with respect to the j−th variable in the point x the

∂j f (x) := De j f (x).

Remark 3.1.4. Partial derivative ∂j is nothing but an ordinary derivatives w.r.t xj considering all othervariables xi i , j as fixed parameters. Indeed

∂j f (x) = limt→0

f((x1, . . . , xj−1, xj, xj+1, . . . , xd) + t(0, . . . ,0,1,0, . . . ,0)

)− f (x1, . . . , xd)

t

= limt→0

f(x1, . . . , xj−1, xj + t, xj+1, . . . , xd

)− f (x1, . . . , xj−1, xj, xj+1, . . . , xd)

tSo, for instance

∂x (y sin x) = y cos x, ∂y (y sin x) = sin x.

37

3.2. Differential

Let’s take the definition of derivative from another side. Notice that, for real functions of real variable,

f ′(x) = limh→0

f (x + h) − f (x)h

, ⇐⇒ limh→0

f (x + h) − f (x) − f ′(x)hh

= 0,

that isf (x + h) − f (x) − f ′(x)h = o(h).

If f : D ⊂ R −→ R, f ′(x) is a number and f ′(x)h is the algebraic product between f ′(x) and h. If nowf : D ⊂ Rd −→ Rm we expect that

i) f ′(x)h is of the same nature of f (x + h) and f (x), that is f ′(x)hRm (this because the quantityf (x + h) − f (x) − f ′(x)h should make sense);

ii) f ′(x)h depends linearly by h.In other words, f ′(x) should be something that works linearly on vectors h ∈ Rd to vectors of f ′(x)hRm.The natural object for f ′(x) is therefore an m × d matrix. This motivates the

Definition 3.2.1. Let f : D ⊂ Rd −→ Rm, x ∈ Int(D). We say that f is differentiable in x iff thereexists a m × d matrix denoted by f ′(x) and called jacobian matrix such that

(3.2.1) f (x + h) − f (x) − f ′(x)h = o(h),

in the sense that

(3.2.2) limh→0

‖ f (x + h) − f (x) − f ′(x)h‖‖h‖

= 0.

The natural question is: what are the entries of the jacobin matrix?

Proposition 3.2.2. If f is differentiable in x then there exists all the directional derivatives of f in x and

(3.2.3) Dv f (x) = f ′(x)v, ∀v ∈ Rd .

In particular, if the components of f are f = ( f1, . . . , fm) then

(3.2.4) f ′(x) =

∂1 f1(x) ∂2 f1(x) . . . ∂d f1(x)

......

...∂1 fm(x) ∂2 fm(x) . . . ∂d fm(x)

.Proof — Let’s prove the (3.2.3). Fix v , 0. Then, by (3.2.2)

f (x + tv) − ( f (x) + f ′(x)(tv)) = o(tv), =⇒f (x + tv) − f (x)

t= f ′(x)v +

o(tv)t

.

Now, because

limt→0

o(tv)t

= limt→0

‖o(tv)‖|t |

= ‖v‖ limt→0

‖o(tv)‖‖tv‖

= ‖v‖ limh→0

‖o(h)‖‖h‖

= 0,

we conclude thatlimt→0

f (x0 + tv) − f (x0)

t= f ′(x0)v.

38

Let’s prove now the (3.2.4): if we call f ′(x0) = [ai j], it is well known by Linear Algebra that

f ′(x0)ej

gives the j−th column of the matrix f ′(x0). So the element ai j of f ′(x0) is obtained by taking the i−th componentof the vector f ′(x0)ej . But: by (3.2.3) we have

f ′(x0)ej = De j f (x0) = ∂j f (x0) =(∂j f1(x0), ∂j f2(x0), . . . , ∂j fm(x0)

),

hence the i−th component is ∂j fi(x0), and this proves (3.2.4).

Remark 3.2.3. In particular, differentiable =⇒ directionally derivable =⇒ partially derivable.

Two important cases are• f : D ⊂ Rd −→ R: in this case f ′(x) is a 1 × d matrix, precisely

f ′(x) = [∂1 f (x) ∂2 f (x) . . . ∂d f (x)] =: ∇ f (x),

called gradient of f in x. In this casef ′(x)h = ∇ f (x) · h,

where we denoted by · the scalar product of vectors in Rd.• γ : [a, b] ⊂ R −→ Rd: in this case γ′(t) is a d × 1 matrix, precisely

γ′(t) =

γ′1(t)...

γ′d(t)

.Example 3.2.4. Discuss the differentiability at (0,0) of

f (x, y) :=

x2y2

x2+y2 , if (x, y) , (0,0),

0, if (x, y) = (0,0).

Sol. — We know that the candidate for f ′(0,0, ) = ∇ f (0,0) if it exists. Notice that we cannot simply computepartial derivatives and evaluate in (0,0) because, for instance,

∂x f (x, y) = ∂xx2y2

x2 + y2 =2xy2(x2 + y2) − x2y22x

(x2 + y2)2=

2xy4

(x2 + y2)2,

is of course not defined in (0,0). In this case we have to proceed directly in the computation of ∂x f (0,0), that is

∂x f (0,0) = D(1,0) f (0,0) = limt→0

f ((0,0) + t(1,0)) − f (0,0)t

= limt→0

f (t,0)t= lim

t→0

0t= 0,

and similarly ∂y f (0,0) = 0. We deduce ∇ f (0,0) = (0,0). What it remains to do? We have to check that

f (02 + h) − f (02) − ∇ f (02) · h = o(h), ⇐⇒ limh→02

‖ f (02 + h) − f (02) − ∇ f (02) · h‖‖h‖

= 0.

Now: call h = (u, v):

f (02 + h) − f (02) − ∇ f (02) · h = f (u, v) − 0 − (0,0) · (u, v) = f (u, v).

39

We have therefore to prove that

lim(u,v)→(0,0)

f (u, v)‖(u, v)‖

= 0.

This is a limit in R2 that we will compute by using the methods of previous chapter. Notice that

f (u, v)‖(u, v)‖

=

u2v2

u2+v2√

u2 + v2=

u2v2

(u2 + v2)3/2u=ρ cos θ, v=ρ sin θ

=ρ4(cos θ)2(sin θ)2

ρ3 = ρ(cos θ)2(sin θ)2,

hence �� f (u, v)‖(u, v)‖

�� 6 ρ −→ 0, as ρ −→ 0.

This finishes the exercise and prove that f is differentiable in 02 and f ′(02) = (0,0).

To check differentiability by using the definition is long and complex. A useful test is the following

Theorem 3.2.5 (total differential theorem). Let f = ( f1, . . . , fm) : D ⊂ Rd −→ Rm, D open. If

∂j fi ∈ C (D), ∀i, j, =⇒ ∃ f is differentiable in any x ∈ D

A function f fulfilling this hypothesis is called a C 1(D) function.

Abovewe proved that to be differentiable implies that all the directional derivatives exists. Differentiabilityis actually a stronger concept. This follows as by product of the following

Proposition 3.2.6. If f is differentiable at x it is therein continuous.Proof — By (3.2.2) we have f (y) = f (x) + f ′(x)(y − x) + o(y − x) −→ f (x), as y −→ x.

The rules of calculus of differentials basically are the same of those of ordinary calculus. For instance

( f + g)′(x) = f ′(x) + g′(x).

if f ,g are differentiable at x. Similarly it holds the important

Theorem 3.2.7 (chain rule). Let f : D ⊂ Rd −→ Rm, g : E ⊂ Rm −→ Rk , x ∈ Int(D) andf (x) ∈ Int(E) such that ∃ f ′(x) and ∃ g′( f (x)). Then

(3.2.5) ∃ (g ◦ f )′(x) = g′( f (x)) f ′(x).

A special important case of (3.2.5) is the following: suppose we want to computeddtg(γ(t)), where γ : I ⊂ R −→ Rm, g : E ⊂ Rm −→ R.

Assuming all the hypotheses fulfilled we have

(3.2.6)ddtg(γ(t)) = g′(γ(t))γ′(t) = [∂1g(γ(t)) . . . ∂mg(γ(t))]

γ′1(t)...

γ′m(t)

= ∇g(γ(t)) · γ′(t).

This is also called total derivative of g along γ.

40

3.3. Extrema

In this section we study the problem of finding min/max points of a function. Let’s introduce first theimportant

Definition 3.3.1. Let f : D ⊂ Rd −→ R. We say that a point x0 ∈ D is• global maximum (minimum) of f on D if f (x) 6 f (x0) ( f (x) > f (x0)) for any x ∈ D;• local maximum (minimum) of f on D if there exists a neighborhood Ux0 ⊂ D of x0 such that

f (x) 6 f (x0) ( f (x) > f (x0)) for any x ∈ Ux0 .

The goal of this section is to show how differential calculus can be used to determine local min/maxpoints. Let’s start by the extension of a well known property of Calculus in one variable:

Theorem 3.3.2 (Fermat). Let f : D ⊂ Rd −→ R and x0 ∈ Int(D) be a local min/max. If f isdifferentiable at x0 then ∇ f (x0) = 0.Proof — The idea is quite easy: if x0 is (for instance) a local maximum for f , it is also local maximum on anysection. Let’s translate formally this idea: assuming, for instance, that x0 is a local maximum for f on D,

∃ r > 0, : f (x) 6 f (x0), ∀x ∈ B(x0,r] ∩ D.

Because x0 ∈ Int(D) we may assume directly that B(x0,r] ⊂ D. Now, consider f on a straight line passing by x0with direction v: f (x0 + tv). Once x0 + tv ∈ B(x0,r] (and this happens iff |t | 6 r

‖v ‖ ) we will have

f (x0 + tv) 6 f (x0), ∀t ∈ I0 :=[−

r‖v‖

,r‖v‖

].

This condition says that t 7−→ f (x0 + tv) has a maximum at t = 0. By classical Fermat thm for one variable case,we deduce that

0 =ddr

f (x0 + tv)��t=0

(3.2.6)= ∇ f (x0) · v.

This must be true for any v ∈ Rd , that is ∇ f (x0) · v = 0, ∀v ∈ Rd . This means that ∇ f (x0) is orthogonal to anyvector v of Rd . Hence ∇ f (x0) = 0.

Be careful! First: the condition ∇ f (x0) = 0 works only if x0 ∈ Int(D).

Example 3.3.3. Let f (x, y) = x2 + y2 on D = {x2 + y2 6 1}. Clearly minD f = 0 and the minimum isattained at (0,0), maxD f = 1 and every point on {x2 + y2 = 1} is a maximum point for f on D. Of course thesepoints are not in Int(D) = {x2 + y2 < 1} and being ∇ f (x, y) = (2x,2y) we see that ∇ f (x, y) = 02 iff (x, y) = (0,0),so ∇ f , 0 at every (x, y) ∈ {x2 + y2 = 1}.

Second: the condition ∇ f = 0 not necessarily identifies min/max.

Example 3.3.4. Let f (x, y) = x2 − y2 on D = R2. Clearly, ∇ f (x, y) = (2x,−2y), therefore ∇ f (0,0) = (0,0).However (0,0) is not an extremum because if we section f along the x axis we get f (x,0) = x2, and this says that(0,0) is a minimum (even global) for this section, whereas when we take the y axis section f (0, y) = −y2 we havethat (0,0) is a global maximum for the section. Again, taking f (x, x) = 0 we get that the function is constant!

41

There’s no surprise with the previous example because the same happens for functions of real variable.For instance, f (x) = x2 on [−1,1] has max at x = ±1 but f ′(±1) , 0 and f (x) = x3 on R has no min/maxbut f ′(0) = 0. In particular, points where ∇ f = 0 are not necessarily min/max, but we will give to thema name:

Definition 3.3.5. If ∇ f (x0) = 0 we say that x0 is a stationary point for f .

Example 3.3.6. Letf (x, y) = xyex

2+y2, (x, y) ∈ D := {(x, y) ∈ R2 : x > 0, y > 0, x2 + y2 6 1}.

Draw D. What can you say about D: is open? closed? compact? connected? Show that f admits globalextrema on D and find these points. Finally, determine f (D).

ÿ

x

y

Figure 1. The boundary of D is colored in thick orange.

Sol. — Being D defined by large inequalities it is closed. It is not open because the points on the boundary belongsto D. Moreover, D is a subset of unit disk, therefore is bounded, hence is compact. A picture of D is easy.

Existence. Because f is continuous and D compact, Weierstrass’s thm says that f has global min and max.

Determination. Because D is not open the argument (which is, by the way, a standard argument) requires somecare. Assume (x, y) be a min/max. We have two cases: either (x, y) ∈ Int(D) (hence it is necessarily a stationarypoint according to Fermat Theorem) or (x, y) ∈ ∂D (then it is not necessarily a stationary point and we have todiscuss directly).

If (x, y) ∈ Int(D) then, by Fermat thm, ∇ f (x, y) = 0. We have

∇ f (x, y) =(ex

2+y2(y + 2x2y), ex

2+y2(x + 2y2x)

)= 0, ⇐⇒

y(1 + 2x2) = 0,

x(1 + 2y2) = 0,⇐⇒ x = y = 0.

But this means that there aren’t stationary points of f in Int(D). In particular, the extrema belongs to ∂D.Hence, (x, y) ∈ ∂D. We have to proceed by direct inspection. First of all∂D = {(x,0) : 0 6 x 6 1} ∪ {(0, y) : 0 6 y 6 1} ∪

{(x, y) : x2 + y2 = 1, x > 0, y > 0

}=: A ∪ B ∪ C.

On A we have f (x,0) = 0, so f is constant; on B we have the same f (0, y) = 0. Let see what happens on C. It’sbetter to describe C in the way

C ={(cos θ, sin θ) : θ ∈

[0,π

2

]}.

In this wayf (cos θ, sin θ) = (cos θ)(sin θ)e1 =

e2

sin(2θ),

42

and clearly this quantity is maximum as 2θ = π2 , that is θ =

π4 , that is in the point

1√2(1,1), whereas the minimum is

0 (as θ = 0, π2 , corresponding to the points (1,0) and (0,1)). In conclusion: on A and B f is constant 0 whereas onC the minimum is 0 (taken on (1,0) ∈ A and (0,1) ∈ B) whereas the maximum is e

2 reached in the point(

1√2, 1√

2

).

The moral is:

minD

f = 0, minimum points: A ∪ B, maxD

f =e2, maximum points:

{(1√

2,

1√

2

)}.

Finally: because D is connected, f (D) is connected in R, hence interval. Of course, by what we have said before,f (D) = [0, e2 ].

Example 3.3.7. Find the eventual global extrema of the function

f : D = B(0,1] ⊂ R2 −→ R, f (x, y) :=√|x + y |e−(x

2+y2).

Sol. — Existence. Notice first that f ∈ C (D) and clearly D is compact. By Weierstrass’s Thm f admits globalmin/max.

Determination. Let (x, y) ∈ D be an extreme point. We have the following alternative: either (x, y) ∈ Int(D) or(x, y) ∈ ∂D.

Case (x, y) ∈ Int(D). If f is differentiable at (x, y) then ∇ f = 0. We have to specify if f is differentiable because fis not differentiable where x + y = 0 (because of the term |x + y |). On that points however f (x, y) ≡ 0. If x + y , 0we need ∇ f = 0. Let’s compute the gradient.

∂x f = sgn(x+y)2√|x+y |

e−(x2+y2) +

√|x + y |e−(x

2+y2)(−2x) = e−(x2+y2)

2√|x+y |

(sgn(x + y) − 4x |x + y |) ,

∂y f = e−(x2+y2)

2√|x+y |

(sgn(x + y) − 4y |x + y |) .

Then ∇ f (x, y) = 02 iffsgn(x + y) − 4x |x + y | = 0,

sgn(x + y) − 4y |x + y | = 0,⇐⇒

sgn(x + y)(1 − 4x(x + y)) = 0,

sgn(x + y)(1 − 4y(x + y)) = 0,⇐⇒

x(x + y) = 1

4 ,

y(x + y) = 14 .

By this, easily x, y , 0 and xy = 1, that is x = y, and plugging again into the system be get x(2x) = 1

4 , that isx2 = 1

2 , x = ± 1√2. Therefore we find(

1√

2,

1√

2

),

(−

1√

2,−

1√

2

)∈ Int(D)\{y = −x}.

Notice that

f(

1√

2,

1√

2

)=

√2√

2e−

12 =

4√2√

e.

Case (x, y) ∈ ∂D = {x2 + y2 = 1} ∪ {y = −x}. On {y = −x} we have f (x,−x) = 0, and because f > 0 clearly forevery (x, y) ∈ R2, the points {y = −x} are surely global minimum points; on {x2 + y2 = 1} it is convenient to usethe standard parametrization: we have

f (cos θ, sin θ) =√| cos θ + sin θ |e−1.

43

It is easy to check that cos θ + sin θ gets its maximum value as θ = π4 ,

34π. Therefore

maxx2+y2=1

f (x, y) =

√2√

2e−1 =

4√2e−1 <4√2e−1/2 = f

(1√

2,

1√

2

)= f

(−

1√

2,−

1√

2

).

By this follows that(

1√2, 1√

2

)e

(− 1√

2,− 1√

2

)are global maximum.

3.4. ExercisesExercise 3.4.1. Compute the following directional derivatives (if they exists):

1. D(√

3,1) log(1 + x2y2), at (1,1). 2. D(2,2) arctan(x + y), at (1,0). 3. D(1,1)x2y

|x | + y2 , at (0,0).

4. D(−1,1)xy

x2 + y4 , at (0,0). 5. D(−1,−2)y(ex − 1)√

x2 + y2, at (0,0).

Exercise 3.4.2. For each of the following functions say if a) is continuous at point (0,0); ii) there exist ∂x f (0,0),∂y f (0,0); iii) is differentiable in (0,0).

1. f (x, y) :=

x3

x2+y2 , (x, y) , 02,

0, (x, y) = 02.

2. f (x, y) :=

x2+y2

|x |+ |y | , (x, y) , 02,

0, (x, y) = 02.

3. f (x, y) :=

x2y3

(x2+y2)2, (x, y) , 02,

0, (x, y) = 02.

4. f (x, y) :=

x2y

x2+y2 + x − y, (x, y) , 02,

0, (x, y) = 02.

Exercise 3.4.3. Show that the function f (x, y) = x√

x2 + y2, (x, y) ∈ R2 is differentiable on R2.

Exercise 3.4.4. Determine the stationary points of each of the following functions:

1. f (x, y) = xy(x + 1). 2. f (x, y) = x2 + y2 + xy. 3. f (x, y) = x3 + y3 + 2x2 + 2y2 + x + y.

4. f (x, y) = xey + yex . 5. f (x, y, z) = (x3 − 3x − y2)z2 + z3.

Exercise 3.4.5. For each of the following functions a) find the stationary points, b) find eventual min/max on thedomain, c) find the image of the domain.

(1) f (x, y) = x4 + y4 − xy, on D = R2.(2) f (x, y) = x

((log x)2 + y2) on D =]0,+∞[×R.

(3) f (x, y) = xy(x + y), on D = R2.(4) f (x, y, z) = x2 + 3y2 + 2z2 − 2xy + 2xz on D = R3.(5) f (x, y, z) = x4 + y4 + z4 − xyz, on D = R3.

Exercise 3.4.6. Letf (x, y) = (x2 + y2)2 − 3x2y, (x, y) ∈ R2.

i) Compute (if it exists) lim(x,y)→∞2 f (x, y). ii) Find stationary points of f . iii) Find eventual global min/max of fon R2 and find f (R2).

44

Exercise 3.4.7. Letf (x, y, z) := (x2 + y2 + z2)2 − xyz, (x, y, z) ∈ R3.

i) Show that lim(x,y,z)→∞3 f (x, y, z) = +∞. ii) Find stationary points of f . iii) Show that f has global minimum onR3 and find f (R3).

Exercise 3.4.8. Let f (x, y) := x2(1 − y) on D := {(x, y) ∈ R2 : x2 + |y | 6 4}. Study the sign of f , determine itseventual stationary points on D and find min/max of f on D.

Exercise 3.4.9. Find the extrema of f (x, y) := xye−xy on D = {(x, y) ∈ R2 : 1 6 x 6 4, y > 0, |xy | 6 1}.

Exercise 3.4.10. Find the value of the parameter λ ∈ R such that the function f (x, y) := x2 + λy2 − 4x + 2y has astationary point in (2,−1). What kind of point is this?

Exercise 3.4.11. Let f (x, y) := x2 (y2 − (x − 1)2

), (x, y) ∈ R2. i) Does it exists lim(x,y)→∞2 f (x, y)? If yes,

compute it. ii) Find and classify the stationary points of f on R2. iii) What about extrema of f on R2? Determinef (R2). iv) Show that f has min/max on D := {(x, y) ∈ R2 : y 6 0, 0 6 x 6 y + 1} and find them. What is f (D)?

Exercise 3.4.12. Consider the function f (x, y) := x4 + y4 − 8(x2 + y2) on R2. i) Compute lim(x,y)→∞2 f (x, y).ii) Find and classify the stationary points of f . What can you say about global extreme points of f ? What aboutf (R2)? iii) Find the extreme points of f on the domain D := {(x, y) ∈ R2 : x2 + y2 6 9}.

Exercise 3.4.13. Consider the function

f (x, y) :=

x5y2

(x4+y2)2, (x, y) , 02,

0, (x, y) = 02.

i) Say if f is continuos, differentiable in 02 (and in this case compute ∇ f (02)). ii) Find the eventual stationarypoints of f on R2 and discuss their nature. Does f has extreme points on R2? iii) Show that f has min/max on thedomain D = {(x, y) ∈ R2 : |x | 6 1, |y | 6 1} and find them.

Exercise 3.4.14. Let f be the function defined as

f (x, y) :=

xyexy

x2+y2 , (x, y) ∈ R2\{02},

0, (x, y) = 02.

i) Say if f is continuous and differentiable at 02. ii) Does it exists lim(x,y)→∞2 f (x, y)? (in the case affirmative,what is the value?). Is f bounded on R2? iii) Show that f has min/max on {(x, y) ∈ R2 : x2 + y2 6 1} and findthem.

CHAPTER 4

Constrained Optimization

Many applied questions can be formalized as the maximization/minimization of a certain quantity(function) f of several variables over certain constraints on the variables. For instance: consider theproblem to find the parallelepiped with maximum volume among those with fixed surface S. This meansto determine

maxx,y,z>0 : 2(xy+yz+xz)=S

xyz.

A general form for this problem isfindmax/min f (x1, . . . , xd) subject g1(x1, . . . , xd) = 0, . . . ,gk(x1, . . . , xd) = 0.

Themethod developed in the previous Chapter, based on the individuation of the stationary points, doesn’twork in this context. Indeed, the set of points

M := {(x1, . . . , xd) ∈ Rd : g1(x1, . . . , xd) = 0, . . . ,gk(x1, . . . , xd) = 0},has no interior points in general. This means that conditions of Fermat’s Thm are never fulfilled here,hence min/max are not necessarily stationary points for f . To tackle this problem, let’s consider asimplified version of the initial isoperimetric problem, that is let’s consider the problem to find, amongall the rectangles with fixed perimeter S, those with maximum area. Formally, we want to determine

max(x,y)∈]0,+∞[ : 2(x+y)=S

xy.

In this case, of course, we can reduce the problem to a well known one just by noticing that, by theconstraint 2(x + y) = S we have, in particular, y = S

2 − x. Hence we have to maximize

xy = x(

S2− x

).

This is now a function of one variable that we’ve to maximize for 0 6 x 6 S2 . To do this we can use the

tools of one variable calculus. Trying to catch the moral, we could think that if we have to solvemax

g(x,y)=0f (x, y),

by the equation g(x, y) = 0 we can "extract" for instance y = ϕ(x). Then the problem can be reduced tomaxx

f (x, ϕ(x)).

This last, differently from the former, looks like an "unconstrained maximization" to which we can applythe usual tools of calculus. Of course, to make real this idea, there’re two main points to solve:

• first, is it always possible say that g(x, y) = 0 iff y = ϕ(x)?45

46

• second, admitting ϕ exists, how this can be used to maximize f (x, ϕ(x))?The answer to the first question is given by the Dini’s theorem, the second is the so called Lagrange’smultipliers theorem. For pedagogical reasons, we will first see these results in the simplest (but nontrivial) case when f = f (x, y) and g = g(x, y), then we will extend to the general case.

4.1. Implicit functions: scalar case

Let’s consider a set of the formM := {(x, y) ∈ R2 : g(x, y) = 0}

where g : R2 −→ R. Excluding degenerate cases, we expect that M be a curve in R2. This curve shouldbe actually a graph of a function because, at least intuitively, through the equation we should able toexpress one of the two coordinates as function of the other. For instance, if the equation is x2+ y2−1 = 0(that is x2 + y2 = 1) we can see that x = ±

√1 − y2 as well as y = ±

√1 − x2.

8gHx,yL=0<

8gHx,yL=0<ÝUIx0 ,y0 M =8y=jHxL<

UIx0 ,y0 MHx0 ,y0 L

j

8gHx,yL=0<ÝUH-1,0L =8x=ΨHyLL<UH-1,0L

H-1,0LΨ

If we fix a point (x0, y0) ∈M , then in a neighborhood of (x0, y0) the set M is precisely the graph of justone function of type y = ϕ(x) or x = ψ(y). Notice that for all points except (±1,0) this graph could bein the form y = ϕ(x) but for points (±1,0) it is impossible to represents the set of the solution (namelythe circumference) as a graph of this type. It is, however, possible to express the set as a graph of typex = ψ(y).

Suppose that the set M = {g(x, y) = 0} be, locally, the graph of a function of type y = ϕ(x), that is{g(x, y) = 0} ∩U(x0,y0) = {y = ϕ(x)}.

Assuming that ϕ be also a regular function,

(4.1.1) g(x, ϕ(x)) ≡ 0, =⇒ 0 ≡ddx

g(x, ϕ(x)) = ∂xg(x, ϕ(x)) + ∂yg(x, ϕ(x))ϕ′(x).

If ∂yg(x, ϕ(x)) , 0 to get

ϕ′(x) = −∂xg(x, ϕ(x))∂yg(x, ϕ(x))

.

In particular, setting x = x0, being y0 = ϕ(x0) we would have ϕ′(x0) = −∂xg(x0,y0)∂yg(x0,y0)

.

UHx0 ,y0L

Ux0

Hx0,y0L

y=jHxL

gHx,yL=0

x0

x

y

47

The condition ∂yg(x0, y0) , 0 turns out to be the right condition for the existence of the function ϕ:

Theorem 4.1.1 (Dini). Let g : D ⊂ R2 −→ R, D open and g ∈ C 1(D). Let (x0, y0) ∈ D such thatg(x0, y0) = 0 and suppose that

(4.1.2) ∂yg(x0, y0) , 0.

There exists then U(x0,y0) neighborhood of (x0, y0) in D, Ux0 neighborhood of x0 in R, ϕ : Ux0 −→ R suchthat

{g = 0} ∩U(x0,y0) ={(x, ϕ(x)) : x ∈ Ux0

}.

Moreover ϕ ∈ C1 and

(4.1.3) ϕ′(x) = −∂xg(x, ϕ(x))∂yg(x, ϕ(x))

, ∀x ∈ Ux0 .

The function ϕ is called implicit function defined by g.

Remark 4.1.2. A similar statement holds to explicit x = ψ(y). The key hypothesis that replaces (4.1.2) is

(4.1.4) ∂xg(x0, y0) , 0.

Then exists ψ : Uy0 ⊂ R −→ R, such that {g = 0} ∩U(x0,y0) = {(ψ(y), y) : y ∈ Uy0}.

Remark 4.1.3. In particular: if ∇g(x0, y0) , (0,0) we have that {g = 0} is locally the graph of somefunction y = ϕ(x) or x = ψ(y) in a neighborhood of (x0, y0).

Remark 4.1.4 (Warning!). Dini Thm gives a sufficient condition to explicit one variable in term ofthe other. A frequent error is to think that the hypothesis (4.1.2) (or (4.1.4)) is sufficient: this isfalse! In other words, a common error is to believe that "if one of (4.1.2) or (4.1.4) is not fulfilled, then is notpossible to explicit one variable in term of the other". Look at the following "stupid" example: let

g(x, y) = (x − y)2.

Of course {g(x, y) = 0} = {y = x} is a global graph of the function y = x or x = y. But

∂yg(x, y) = −2(x − y) ≡ 0, ∀(x, y) ∈ {g = 0},

as well ∂xg(x, y) ≡ 0 for every (x, y) ∈ {g = 0}. Therefore, hypotheses (4.1.2) and (4.1.4) are never fulfilled!

Example 4.1.5. Consider the equation

x3 + y3 − 3xy − 3 = 0.

Show that if (x, y) is a solution of this equation is always possible to explicit at least one between x andy as function of the other variable.Sol. — Define g(x, y) := x3 + y3 − 3xy − 3. If we prove that when (x, y) is a solution then at least one between∂xg(x, y) or ∂yg(x, y) is different from 0 then at least one of (4.1.2) and (4.1.4) is fulfilled: therefore, Dini Thm

48

applies at least in one of the two cases and we are done. To this aim let’s see if there are points on {g = 0} whereboth (4.1.2) and (4.1.4) fail: we have to find solution of

(x, y) ∈ {g = 0},

∂xg(x, y) = 0,

∂yg(x, y) = 0.

⇐⇒

x3 + y3 − 3xy − 3 = 0

3x2 − 3y = 0,

3y2 − 3x = 0.

⇐⇒

x3 + y3 − 3xy − 3 = 0,

x2 = y,

y2 = x.

By two last equations y4 = y, that is y(y3 − 1) = 0 whose solutions are y = 0,1. As y = 0 we have x = 0, and fory = 1 we have x = 1, that is the points (0,0) and (1,1). Now the question is: do they satisfy also the first condition?It is easy to check that the answer is no!

4.2. Lagrange multipliers: scalar case

Let’s now consider the problemminM/max

Mf (x, y), where M := {g(x, y) = 0}.

Our goal here is to find an analogous of condition ∇ f = 0 for constrained min/max points. Assume that(x0, y0) be a min/max for f on M and assume that, in a neighborhood of (x0, y0), M be the graph ofsome regular function. According to Dini’s Theorem and, in particular, to the Remark 4.1.3, a sufficientcondition for this is

∇g(x0, y0) := (∂xg(x0, y0), ∂yg(x0, y0)) , 02.

Definition 4.2.1. We say that g is submersive at (x0, y0) if ∇g(x0, y0) , 02.

In this setting we have

Theorem 4.2.2 (Lagrange). Assume that f ∈ C 1(D), D ⊃ M := {g = 0}, g ∈ C 1(R2), g submersiveat (x0, y0). If (x0, y0) ∈M is a local min/max for f on M then(4.2.1) ∃λ ∈ R : ∇ f (x0, y0) = λ∇g(x0, y0).

Points (x0, y0) where (4.2.1) holds are called constrained stationary points.Proof — Suppose that, for instance, ∂yg(x0, y0) , 0 (the other case is treated similarly). Then, by Dini’s theorem,there exists a neighborhood U of (x0, y0) and a function y = ϕ(x) such that

M ∩U = {(x, ϕ(x)) : x ∈ Ix0 }.

Now, because (x0, y0) is a local minimum for f on M (same argument in the case of a maximum),f (x, ϕ(x)) > f (x0, ϕ(x0)), ∀x ∈ Ix0,

that is the auxiliary function h(x) := f (x, ϕ(x)) has a minimum at x0. Therefore, by the one variable Fermattheorem,

h′(x0) = 0.But

h′(x) =ddx

f (x, ϕ(x)) = ∂x f (x, ϕ(x)) + ∂y f (x, ϕ(x))ϕ′(x),

hence0 = ∂x f (x0, y0) + ∂y f (x0, y0)ϕ

′(x0).

49

Now, recalling that for the implicit function ϕ we have the (4.1.3) we deduce

0 = ∂x f (x0, y0) − ∂y f (x0, y0)∂xg(x0, y0)

∂yg(x0, y0),

that is

∂x f (x0, y0)∂yg(x0, y0) − ∂y f (x0, y0)∂xg(x0, y0) = 0, ⇐⇒ ∇ f (x0, y0) ⊥(∂yg(x0, y0),−∂xg(x0, y0)

).

But then∇ f (x0, y0) ∝ (∂xg(x0, y0), ∂yg(x0, y0)) = ∇g(x0, y0).

In practice, to determine constrained min/max we can proceed as follows. First we make sure if theyexists (by a Weierstrass like argument). Then we look for constrained stationary points of f on M . Themin/max is among them. Finally, we could just evaluate f on these points: those where f is minimumare the mins, those where f is maximum are the maxs.

Example 4.2.3. Find points of the ellipse x2 + 2y2 − xy = 9 at min/max distance to the origin.Sol. — Let M := {x2 + 2y2 − xy = 9} = {x2 + 2y2 − xy − 9 = 0} =: {g = 0}. We have to minimize/maximizethe distance to the origin, that is the function

f (x, y) =√

x2 + y2.

Because of the properties of the root, to minimize this function or just x2 + y2 is the same (it produces the samepoints but of course not the same values!) being

√x2 + y2 min/max iff x2 + y2 it is, we replace the previous f with

f (x, y) = x2 + y2,

which is easier to be managed.

Existence: f ∈ C (R2) and M is clearly closed. If we don’t recognize an ellipse immediately (hence we canconclude that M is also bounded) we can easily show easily this: recalling that

xy 612(x2 + y2),

if (x, y) ∈M then

x2 + 2y2 = 9 + xy 6 9 +12(x2 + y2), =⇒

12

x3 +32y2 6 9, =⇒

12

x2,32y2 6 9,

by which x2 6 18 (hence |x | ≤√

18) and y2 6 6 (that is |y | 6√

6). In any case both x, y are bounded hence M isbounded. The conclusion is that, according to Weierstrass theorem, M is compact, hence f admits both min/max.

Determination: by the previous argument we know that min/max points for f exist. Let’s see if we can apply theprevious theorem. We need to check if g is submersive on M . To this aim let’s see where g is not submersive.This happens iff

∇g = 0, ⇐⇒ (2x − y,4y − x) = (0,0), ⇐⇒

2x − y = 0,

4y − x = 0,⇐⇒ x = y = 0.

Therefore g is not submersive at point (0,0) <M , hence g is submersive on M .

50

According to the previous theorem in a min/max point we must have ∇ f ∝ ∇g. Being ∇ f = (2x,2y) thismeans

(2x,2y) = λ(2x − y,4y − x), ⇐⇒

2x = λ(2x − y),

2y = λ(4y − x)Of course (0,0) is a solution of the system, but because (0,0) < M it cannot be considered as candidate to be enextrema for f . Are there non trivial solutions? Notice that if x = 0 (or y = 0) then necessarily y = 0 (x = 0).Hence we can assume x, y , 0: in such case by dividing the two equations we get

2x2y=

2x − y

4y − x, ⇐⇒ x(4y − x) − (2x − y)y = 0, ⇐⇒ 2xy + y2 − x2 = 0.

This can be rewritten as

(x + y)2 − 2x2 = 0, ⇐⇒ (x + y)2 = 2x2, ⇐⇒ x + y = ±√

2x, ⇐⇒ y = (±√

2 − 1)x.

Therefore we have points (x, (±√

2 − 1)x). Of course we have to look at those of them that belongs to M :

(x, (√

2− 1)x) ∈M , ⇐⇒ x2 + 2(√

2− 1)2x2 − (√

2− 1)x2 = 9, ⇐⇒ (8− 5√

2)x2 = 9, ⇐⇒ x = ±3√

8 − 5√

2.

This produces points(± 3√

8−5√

2,± 3(

√2−1)√

8−5√

2

)(same sign, 2 points). Similarly

(x, (−√

2−1)x) ∈M , ⇐⇒ x2+2(−√

2−1)2x2−(−√

2−1)x2 = 9, ⇐⇒ (8+5√

2)x2 = 9, ⇐⇒ x = ±3√

8 + 5√

2.

This produces points(± 3√

8+5√

2,± 3(−

√2−1)√

8+5√

2

)(same sign, two points). Now, being

f

(±

3√8 − 5

√2,±

3(√

2 − 1)√8 − 5

√2

)=

36 − 18√

28 − 5

√2

> f

(±

3√8 + 5

√2,±

3(−√

2 − 1)√8 + 5

√2

)=

36 + 18√

28 + 5

√2

we have that the first points are max for f (hence point of M at max distance from 02), the latter are min.

4.3. Dini Theorem: case of systems

A natural extension of the classical Dini’s theorem is the following. Consider a set defined as

M := {(x1, . . . , xd) ∈ Rd : g1(x1, . . . , xd) = 0, . . . ,gn(x1, . . . , xd) = 0}.

The intuitive idea is that, under good conditions, it should be possible to represent points (x1, . . . , xd) ∈Min a neighborhood of a certain (ξ1, . . . , ξd) ∈M by expressing n coordinates in function of the remainingd − n, as for instance

(x1, . . . , xd) ∈M , ⇐⇒

x1 = ψ1(xn+1, . . . , xd),...xn = ψn(xn+1, . . . , xd).

We can see this under the same form of the Dini’s Thm. Call g = (g1, . . . ,gn) : Rd −→ Rn, and write

x = (x1, . . . , xn), y = (xn+1, . . . , xd),

51

in such a way that (x1, . . . , xd) ∈M iff g(x, y) = 0. We set also x0 := (ξ1, . . . , ξn) and y0 := (ξn+1, . . . , ξd).We look then for a function x = ψ(y) such that

{g = 0} ∩U(x0,y0) = {(ψ(y), y)}.

Such ψ would be a function of y ∈ Rd−n into Rn and should fulfill

g(ψ(y), y) ≡ 0.

Deriving this identity with the chain rule

∂xg(ψ(y), y)ψ′(y) + ∂yg(ψ(y), y) = 0.

Here ∂xg denotes the Jacobian matrix of g respect to x = (x1, . . . , xn) while ∂yg is the same respect toy = (xn+1, . . . , xd),

∂xg =

∂1g1 ∂2g1 . . . ∂ng1∂1g2 ∂2g2 . . . ∂ng2...

...∂1gn ∂2gn . . . ∂ngn

, ∂yg =

∂n+1g1 ∂n+2g1 . . . ∂dg1∂n+1g2 ∂n+2g2 . . . ∂dg2...

...∂n+1gn ∂n+2gn . . . ∂dgn

.

Thereforeψ ′(y) = −[∂xg(ψ(y), y)]

−1∂yg(ψ(y), y),

provided ∂xg(ψ(y), y) be invertible. As y = y0, ψ(y0) = x0 and we have ∂xg(x0, y0) needs to beinvertible. As for the Dini’s Theorem this turns out to be the appropriate condition to ask:

Theorem 4.3.1. Let g = (g1, . . . ,gn) : D ⊂ Rd −→ Rn be a C 1 function and (x0, y0) ∈ Rd be such that

g(x0, y0) = 0. Suppose thatdet ∂xg(x0, y0) , 0,

There exists then an implicit function ψ ∈ C 1 and a neighborhood U(x0,y0) of (x0, y0) such that

{g = 0} ∩U(x0,y0) = {(ψ(y), y)}.

Moreover

(4.3.1) ψ ′(y) = − [∂xg(ψ(y), y)]−1 ∂yg(ψ(y), y).

Example 4.3.2. Show that the system x3 − 3xy2 + z3 + 1 = 0,

x − 2y2 − 3z2 + 4 = 0,is a graph of (y, z) as function of x in a neighborhood of the point (x, y, z) = (1,1,1). Compute y′(1).Sol. — Easily we see that (x, y, z) = (1,1,1) is a solution. The problem asks to express (y, z) as function of x in aneighborhood of (1,1,1): this is possible, according to Dini’s thm, if the jacobian ∂(y,z)g(1,1,1) is invertible, whereof course g(x, y, z) = (x3 − 3xy2 + z3 + 1, x − 2y2 − 3z2 + 4). We have

∂(y,z)g(x, y, z) =−6xy 3z2

−4y −6z

, =⇒ ∂(y,z)g(1,1,1) =−6 3

−4 −6

,

52

clearly invertible (its determinant is 36 + 12 = 48). So the first requirement is fulfilled. For the second, instead touse the (4.3.1) we proceed directly: let’s derive the two equations considering y = y(x) and z = z(x). We get

(x3 − 3xy2 + z3 + 1)′ = 0,

(x − 2y2 − 3z2 + 4)′ = 0,t, ⇐⇒

3x2 − 3(y2 + 2xyy′) + 3z2z′ = 0,

1 − 4yy′ − 6zz′ = 0.

Now, replacing (x, y, z) = (1,1,1) we obtain3 − 3(1 + 2y′(1)) + 3z′(1) = 0,

1 − 4y′(1) − 6z′(1) = 0,⇐⇒

−2y′(1) + z′(1) = 0,

4y′(1) + 6z′(1) = 1,⇐⇒ y′(1) =

116, z′(1) =

18.

4.4. Lagrange Multipliers: general case

Let’s now consider the problem

minM/max

Mf (x1, . . . , xd), on M = {(x1, . . . , xd) ∈ Rd : g1(x1, . . . , xd) = 0, . . . ,gn(x1, . . . , xd) = 0}.

The general argument is similar but technically much more involved than that one presented as specialcase above. We will limit to sketch it.

The first step is to discuss when M = {g = 0} is, locally (that is in a neighborhood of each of itspoints), a graph of some function. According to Dini’s Thm 4.3.1, to express locally M as graph where(xi1, xi2, . . . , xin ) are functions of the remaining d − n we need that

det[∂(xi1 ,xi2 ,...,xin )g] = det

∂xi1g1 ∂xi2g1 . . . ∂xin g1∂xi1g2 ∂xi2g2 . . . ∂xin g2...

.... . .

...∂xi1gn ∂xi2gn . . . ∂xin gn

, 0.

The matrix ∂(xi1 ,xi2 ,...,xin )g is the sub-matrix of the Jacobian matrix of g by which we select columnsi1, i2, . . . , in. Now, because it is indifferent which are the i1, . . . , in, we wish just

∃1 6 i1 < i2 < . . . < in 6 d : det[∂(xi1 ,xi2 ,...,xin )g] , 0,

that is at least one of the n × n sub-determinants of the Jacobian of g be , 0. It is well known that this isequivalent to say that

rank[∂g] = n.

Definition 4.4.1. We say that g : Rd −→ Rn is submersive at x if rank[∂g(x)] = n. If g is submersiveat every point of a set S, we say that g is a submersion on S. In particular, if g is submersion onM := {x ∈ Rd : g(x) = 0} we say that M is a differential manifold.

We notice that ∂g is a n× d matrix (with d > n in our setting), hence to say that rank[∂g] = n means alsothat the rank is maximum. With a proof similar to that one seen above it is possible now to prove

53

Theorem 4.4.2 (Lagrange multipliers theorem). Assume that f ∈ C 1(D;R), D ⊃M := {g = 0}, bea differential manifold. Then, if ξ ∈M is a local min/max for f on M we necessarily have

(4.4.1) ∃λ1, . . . , λn ∈ R : ∇ f (ξ) =n∑i=1

λi∇gi(ξ).

Points ξ where (4.4.1) holds are called constrained stationary points.

The (4.4.1) says that ∇ f (ξ) is linearly dependent by ∇g1(ξ), . . . ,∇gn(ξ) or, equivalently,

(4.4.2) rank[∇ f (ξ) ∇g1(ξ) . . .∇gn(ξ)] = n.

Because g is assumed to be submersive on M , rank[∇g1(ξ) . . .∇gn(ξ)] = rank[∂g] = n, Therefore, tocheck the (4.4.2) it is sufficient to check that all the (n + 1) × (n + 1) sub-determinants of the matrix[∇ f ∇g1 · · · ∇gn] vanish.

Example 4.4.3. Let M :={(x, y, z) ∈ R3 : xy + z2 = 1, x2 + y2 = 1

}. i) Show that M is a non empty

differential manifold. ii) Say if M is compact or less. iii) Find points of M at minimum/maximumdistance to the origin.Sol. — i) Let’s check that M , ∅. To this aim let’s look to points of type (x, x, z) ∈ M . Imposing this weget 2x2 = 1, that is x = ± 1√

2. By the first, then, x2 + z2 = 1, that is z2 = 1 − x2 = 1 − 1

2 =12 , i.e. z = ± 1√

2.

Therefore (± 1√2,± 1√

2,± 1√

2) ∈ M (all combinations of sign provided sign of the first two coordinates are equal).

Let g : R3 −→ R2, g(x, y, z) := (xy + z2 − 1, x2 + y2 − 1). Clearly g ∈ C 1 and M = Z(g). Let’s find points whereg is not submersive. This means

rank g′(x, y, z) < 2, ⇐⇒ rank

y x 2z

2x 2y 0

< 2, ⇐⇒

2(y2 − x2) = 0,−4xz = 0,−4yz = 0.

Now, this produces the two casesx = 0,y2 = 0, ⇐⇒ y = 0,z ∈ R

or{

z = 0,x2 − y2 = 0, ⇐⇒ y = x, ∨ y = −x.

Therefore, g is not submersive at points (0,0, z), z ∈ R and (x, x,0), (x,−x,0), x ∈ R. Which of them belongs toM ? Clearly (0,0, z) <M for any z ∈ R; moreover,

(x, x,0) ∈M , ⇐⇒

{x2 = 1,2x2 = 1, (impossible), (x,−x,0) ∈M , ⇐⇒

{−x2 = 1,2x2 = 1, (impossible).

Therefore g is submersive on M and consequently M is a differential manifold of dimension 1 in R3.ii) Because M = Z(g) and g ∈ C it follows that M is closed. It is also bounded because, by second constraint,x2 + y2 = 1 we deduce |x |, |y | 6 1, and by the first

z2 = 1 − xy 6 2, =⇒ |z | 6√

2.

iii) We should minimize/maximize the function f (x, y, z) =√

x2 + y2 + z2. Because this is min/max exactly whenthe same happens for f (x, y, z) = x2 + y2 + z2, we use this last function to find extrema. In the previous point we

54

have seen that M is compact, hence min/max exist by Weierstrass Thm. Because f ∈ C 1(R3)we have that extremapoints are stationary points. To find them we use the Lagrange Thm (that can be used by point i)): we have

(x, y, z) ∈M stationary point ⇐⇒ rank

F ′(x, y, z)

∇ f (x, y, z)

= 2, ⇐⇒ det

y x 2z2x 2y 02x 2y 2z

= 0.

Computing the determinant by third column,

2z(2y2 − 2x2) = 0, ⇐⇒ z(y − x)(y + x) = 0.

Candidates are therefore the points (x, y,0), x, y ∈ R, (x, x, z), (x,−x, z), with x, z ∈ R. Now

(x, y,0) ∈M , ⇐⇒

x2 = 1,

x2 + y2 = 1,⇐⇒ (x, y,0) = (±1,0,0).

Similarly

(x, x, z) ∈M , ⇐⇒

x2 + z2 = 1,

2x2 = 1,⇐⇒

(±

1√

2,±

1√

2,±

1√

2

);

(x,−x, z) ∈M , ⇐⇒

−x2 + z2 = 1,

2x2 = 1,⇐⇒

(±

1√

2,∓

1√

2,±

√3√

2

);

It is easy to conclude: (±1,0,0) are the points at min distance,(± 1√

2,∓ 1√

2,±√

3√

2

)are at max distance.

Example 4.4.4. Let M := {(x, y, z) ∈ R3 : x2 + y2 + z4 = 1, x2 − yz = 0}. i) Prove that M is adifferential manifold. ii) Prove that M is compact. iii) Find points of M with maximum quote.Sol. — i) M is defined by constraints g1(x, y, z) := x2 + y2 + z4 − 1, g2(x, y, z) := x2 − yz, clearly C 1(R3). Settingg(x, y, z) := (g1(x, y, z),g2(x, y, z)), we have to check that g is submersive on M . Now

g is not submersive on (x, y, z) ⇐⇒ rank g′(x, y, z) = rank

2x 2y 4z3

2x −z −y

< 2,

and this happens iff all the 2 × 2 sub-determinants of g′(x, y, z) vanish. This means−2xz − 4xy = 0,−2xy − 8xz3 = 0,−2y2 + 4z4 = 0

⇐⇒

x(z + 2y) = 0,x(y + 4z3) = 0,y2 = 2z4,

55

which produces the alternatives

(1){

x = 0,y2 = 2z4,

=⇒ (x, y, z) = (0,±√

2z2, z) ∈M ⇐⇒

2z2 + z4 = 1,

±√

2z3 = 0,=⇒ impossible

(2)

z = −2y,y = −4z3,y2 = 2z4.

⇐⇒

z = −2y,y = 8y3,y2 = 32y4.

=⇒

(2a){

y = 0,z = 0, =⇒ (x,0,0) <M

(2b)

z = −2y,1 = 8y2,1 = 32y2,

=⇒ impossible.

All this means that there aren’t points on M on which g is not submersive.ii) Notice that

M ⊂ {(x, y, z) ∈ R3 : x2 + y2 + z4 = 1},(which isn’t the unit ball!), and because

x2 + y2 + z4 = 1, =⇒ x2 6 1, y2 6 1, z4 6 1, =⇒ |x | 6 1, |y | 6 1, |z | 6 1.

Therefore M is bounded. Moreover M is closed because is the zero set of g continuous, and by this follows thatM is compact.iii) We have to maximize the function f (x, y, z) := z on M . Because f is continuous and M compact, theexistence of global max is assured by Weierstrass Thm. Moreover f ∈ C 1 and M is a differential manifold,therefore maximum points have to be stationary for f on M . By Lagrange multipliers Thm, these are necessarilysuch that

(x, y, z) ∈M , s.t. rank∇ f (x, y, z)∇g1(x, y, z)∇g2(x, y, z)

= 2,

and this happens iff

0 = det∇ f (x, y, z)∇g1(x, y, z)∇g2(x, y, z)

= det

0 0 12x 2y 4z3

2x −z −y

= −2xz − 4yz = −2z(x + 2y).

This produces the following alternatives:

z = 0, ⇐⇒ (x, y, z) = (x, y,0), or x + 2y = 0, ⇐⇒ x = −2y, ⇐⇒ (x, y, z) = (−2y, y, z).

We have to check now which of these points belong to M :

(x, y,0) ∈M , ⇐⇒

{x2 + y2 = 1,x2 = 0, ⇐⇒ (x, y, z) = (0,±1,0).

In the second case

(−2y, y, z) ∈M ⇐⇒

{4y2 + y2 + z4 = 1,4y2 − yz = 0, ⇐⇒

5y2 + z4 = 1,

y(4y − z) = 0.

By this we have the alternatives y = 0 (hence z4 = 1 and we find points (0,0,±1)) or z = 4y which produces{5y2 + 44y4 = 1,z = 4y.

56

Solving the first equation we get

y2 =−5 ±

√25 + 45

2 · 44 , ⇐⇒ y =

√−5 +

√25 + 45

2 · 44 =: y.

Therefore we have a further candidate, the point (−2y, y,4y). The maximum point is therefore between (0,±1,0),(0,0,±1) and (−2y, y,4y), and is simply that one with the maximum z. Because 4y < 1 we have easily that (0,0,1)is the maximum.

Example 4.4.5. A segment of length L is divided into n parts x1, . . . , xn. Find the maximum of x1 · · · xn.Deduce by this the classical inequality

n√

x1 · · · xn 6x1 + . . . + xn

n, ∀x1, . . . , xn > 0.

Sol. — We have to findmax

x1+...+xn=L, x1 ,...,xn>0x1 · · · xn.

First: let’s prove that the maximum exists. Indeed, letM := {x1 + . . . + xn = L : x1, . . . , xn > 0} .

Clearly M is an n− 1 differential manifold defined by a unique constraint g(x1, . . . , xn) = x1 + . . .+ xn − L, clearlysubmersive on all Rn (∇g ≡ (1,1, . . . ,1)). In particular M is closed as zero set of a continuous function (g).Moreover is bounded. Indeed, because

x1, . . . , xn > 0, x1 + x2 + . . . + xn = L, =⇒ 0 < xj < L, ∀ j = 1, . . . ,n.Therefore M is compact and because f (x1, . . . , xn) = x1 · · · xn is of course continuous we have the existence byWeierstrass Thm. Now, let’s find the stationary points of f on M . We have (x1, . . . , xn) ∈M is stationary iff

1 = rank∇ f (x1, . . . , xn)

∇g(x1, . . . , xn)

= rank

x2 · · · xn x1x3 · · · xn · · · x1 · · · xn−1

1 1 · · · 1

,that is iff all the 2 × 2 sub determinants vanish. Choosing column i and j respectively we have

det

x1 · · · xi−1xi+1 · · · xn x1 · · · xj−1xj+1 · · · xn

1 1

= x1 · · · xi−1xi+1 · · · xj−1xj+1 · · · xn(xj − xi).

Therefore, (x1, . . . , xn) ∈M is critic for f on M iffx1 · · · xi−1xi+1 · · · xj−1xj+1 · · · xn(xj − xi) = 0, ∀i , j = 1, . . . ,n.

This produces points where a coordinate is null (hence f = 0) and, if xj > 0 for any j, xi − xj = 0 for all i, j, andthis means that (x1, . . . , xn) = (α,α, . . . , α). Imposing that this belongs to M we find the point ( Ln , . . . ,

Ln ) where

f > 0: therefore this is the maximum! The moral is

maxx1+...+xn=L, x1 ,...,xn>0

x1 · · · xn =(

Ln

)n.

In particular, recalling that x1 + . . . + xn = L, this can be rewritten as

x1 · · · xn 6( x1 + . . . + xn

n

)n, ⇐⇒ n

√x1 · · · xn 6

x1 + . . . + xnn

,

that is just the classical inequality between arithmetic and geometric means.

57

Example 4.4.6. Between all the convex polygons inscribed into a circumference, find those of maximumperimeter.Sol. — Let r > 0 be the radius of the circumference, θ1, . . . , θn the subsequent angles formed by the vertexes ofthe polygon. Then

perimeter = P(θ1, . . . , θn) =

n∑j=1

2r sinθ j

2.

Of course 0 < θ j < 2π and θ1 + · · · + θn = 2π. So, we have to find

maxθ1+...+θn=2π, 0<θ j<2π, j=1,...,n

n∑j=1

2r sinθ j

2.

LetM := {(θ1, . . . , θn) ∈]0,2π[n : θ1 + · · · + θn = 2π} .

Clearly M is and n − 1 dimensional differential manifold in Rn. An argument similar to that one of the previousexample, shows that the maximum exists. Let’s find stationary points of P on M . These fulfills

rank

r cos θ12 · · · r cos θn2

1 · · · 1

= 1, ⇐⇒ r cosθi2= r cos

θ j

2, ∀i, j, ⇐⇒ θi = θ j, ∀i, j .

Therefore, the polygon with maximum perimeter has θ1 = θ2 = . . . = θn =2πn , so it is a regular polygon.

4.5. ExercisesExercise 4.5.1. Determine min/max of f on the set D in the following cases:

i f = x + y, D = {(x, y) : x2 + y2 = 1};ii) f = 2x2 + y2 − x, D = {(x, y) : x2 + y2 = 1};iii) f = xy, D = {(x, y) : x2 + y2 + xy − 1 = 0};iv) f = x2 + 5y2 − 1

2 xy, D = {(x, y) : x2 + 4y2 = 4};v) f = x − 2y + 2z, D = {(x, y, z) : x2 + y2 + z2 = 9};vi) f = z2exy , D = {(x, y, z) : x2 + y2 + z2 = 1}.

Exercise 4.5.2. Let M := {(x, y, z) ∈ R3 : z2 = x2 + y2 + 1, z = 2x2 + y2}. Show that i) M , ∅ is a differentialmanifold of dimension . . . ii) M is compact. iii) M has points of maximum quote: find them.

Exercise 4.5.3. Let M := {(x, y, z) ∈ R3 : z2 = xy + 1}. Show that i) M , ∅ is a differential manifold ofdimension . . . ii) M is not compact. iii) Show that there exists points of M at minimum distance to the origin andfind them.

Exercise 4.5.4. Let M := {(x, y, z) ∈ R3 : (x2 + y2 + z2)2 − xyz = 1}. i) Show that M , ∅ is a differentialmanifold of dimension . . . ii) Say if M is compact or not. iii) Determine, if they exists, points on M at maximumdistance to the origin.

Exercise 4.5.5. Let M :={(x, y, z) ∈ R3 : x2 + y2 − z2 = 0, x2 − y2 = 1

}. i) Show that M , ∅ is a differential

manifold of dimension . . . ii) Say if M is compact or less. iii) Noticed that 03 is not on M , show that exists pointsof M at minimum distance from 03 and find them.

58

Exercise 4.5.6. Let M :={(x, y, z) ∈ R3 : x2 − xy + y2 − z2 = 1, x2 + y2 = 1

}. i) Show that M , ∅ is a

differential manifold of dimension . . . ii) Show that M is compact. iii) Find stationary points of f (x, y, z) = xyzon M . What can you say about the problem to find extrema of f on M ?

Exercise 4.5.7. Let M := {(x, y, z) ∈ R3 : x2 + y2 − z2 = 1}. i) Show that M , ∅ is a differential manifold ofdimension . . . ii) Is M compact? iii) Find points of M at minimum distance from the origin 03.

Exercise 4.5.8. Find the stationary points of f (x, y, z) := xyz, (x, y, z) ∈ R3 on the ellipsoid x2

a2 +y2

b2 +z2

c2 = 1 (herea, b, c > 0). Deduce min/max of f on the ellipsoid.

Exercise 4.5.9. Compute the eventual min/max of f (x, y, z) = xy + yz + zx on the plane x + y + z = 3.

Exercise 4.5.10. Compute the min/max distance of the point (0,1,0) to the following subset of R3:x2 + y2 + z2 = 1,

x2 + y2 = x.

Exercise 4.5.11. Consider the setM :={(x, y, z) ∈ R3 : z = x2 + y2, x + y + z = 0

}. Show thatM is not empty

and is a . . . Find points of M with min/max quote.

Exercise 4.5.12. Let M := {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1, 2z − 3x = 0} and f (x, y, z) := xz. i) Show thatM is non empty and say if it is a differential manifold and what is its dimension. ii) Show that M is compact. iii)Find extrema of f on M .

Exercise 4.5.13. LetM := {(x, y, z) ∈ R3 : 2x2+2y2− z2 = 1, (x− y)2+ z = 2}. i) Show thatM is a differentialmanifold of dimension. . . ii) Show that M is not compact. iii) Find stationary points of f (x, y, z) := z on M .

Exercise 4.5.14. Let

f (x, y, z) :=

√x2 +

y2

4 − 34

+ z2, (x, y, z) ∈ R3.

i) Compute lim(x,y,z)→∞3 f (x, y, z): what can you deduce by this about min/max f ? ii) Find and classify all thestationary points of f on R3. Find, if there exist, min/max of f on R3. What is f (R3)? iii) Let M := {(x, y, z) ∈R3 : f (x, y, z) = 1}. Prove that M is a non empty differential manifold of dimension. . . Is M compact? iv) Showthat there exists points of M at min/max distance to the origin. Find them.

Exercise 4.5.15. Among all the parallelepipedes of sides x, y, z > 0 with fixed total surface find those withmaximum volume.

Exercise 4.5.16. Findmax{xy2z3 : x, y, z > 0, x + y + z = 6}.

Exercise 4.5.17 (?). Let a1, . . . ,ad ∈ R such that a21 + . . . + a2

d> 0. Find

max{x21 + . . . + x2

d : a1x1 + . . . + adxd = 1}.

CHAPTER 5

Integration

In the first Mathematical Analysis course the concept of integral for functions depending on onereal variable has been introduced. This is a central notion in Analysis with several applications: fromthe calculus of areas of plane figures to a fundamental tool in Probability, Physics, Engineering. Werecall that if f = f (x) : [a, b] −→ [0,+∞[, the integral was defined with the following geometricalinterpretation:∫

[a,b]f (x) dx = Area (Trap( f )) , where Trap( f ) := {(x, y) ∈ R2 : x ∈ [a, b], 0 6 y 6 f (x)}.

The definition of Area passes through a formally complex exhaustionmethod based on filling the trapezoidTrap(() f ) with rectangles.

Relevance of integration both in Mathematics and applied disciplines, pushes for the extension of thisoperation to functions of several variables. In this case, however, we don’t have a preferred type of domainsas intervals in case of functions of one real variable. For instance, if f = f (x, y) : E ⊂ R2 −→ [0,+∞[we might expect∫

E

f (x, y) dxdy = Volume (Trap( f )) , where Trap( f ) := {(x, y, z) ∈ R3 : (x, y) ∈ E, 0 6 z 6 f (x, y)}.

Extending this idea, if now f : E ⊂ R3 −→ [0,+∞[, we might expect∫E

f (x, y, z) dxdydz = Hyper-Volume (Trap( f )) ,

where of courseTrap( f ) := {(x, y, z,w) ∈ R4 : (x, y, z) ∈ E, 0 6 w 6 f (x, y, z)}.

Area, Volume and Hyper-Volume are different forms of what we could name measure of a set. As it iscomplicate to construct a solid concept of Area, harder is to define a concept of volume and, more ingeneral, a concept of n−dimensional measure of a subset of Rn. This construction being totally out of ourscope, we will take for granted focussing on the properties of multidimensional integrals. Yet, these are

59

60

too technical to be proved here, thus we will limit to heuristic arguments for their justification, providinghowever proper statements.

5.1. Integral

In the introduction, we mentioned the concept of n−dimensional measure. This is in fact a functionλn that assigns to S ⊂ Rn a number in [0,+∞] (we improperly will treat +∞ as a number). In general, itis not possible to assign a measure to every set. However it is possible to do this on a large class of setsof Rn, called measurable sets.

Theorem 5.1.1. There exists a function

λn : Mn −→ [0,+∞],

whereMn ⊂ P(Rn) is called class (n−dimensional) measurable sets fulfilling the following properties:i) Open and closed sets of Rn are measurable (that is are elements of Mn), in particular ∅ andRn are measurable and

λn(∅) = 0, λn(Rn) = +∞.

Moreover, if a set S differs by an open (closed) set for ameasure 0 set, S is measurable. Precisely,

if ∃E ⊂ Rn open (or closed) : λn ((S\E) ∪ (E\S)) = 0, =⇒ E ∈Mn.

ii) λn factorizes in the sense that

if S = A × B ∈Mn, A ∈Mk, B ∈Mn−k, =⇒ λn(S) = λk(A)λn−k(B).

In particular, λn is coherent with elementary geometry, in the sense that

λn ([a1, b1] × · · · × [an, bn]) = (b1 − a1) · · · (bn − an).

iii) λn is invariant by translations, rotations, reflections and, in general, the following holds true:

(5.1.1) λn(L(S) + v) = | det L |λn(S), ∀L ∈ Mn×n invertible, ∀v ∈ Rn.

(notice that L = In is translation invariance; L =orthogonal matrix LLt = In and v = 0 isrotation invariance).

iv) λn is countably additive, that is if (Sj)j∈N ⊂Mn are disjoint, that is Si ∩ Sj = ∅ if i , j, then

λn©«∞⋃j=1

Sjª®¬ =

∞∑j=1

λn(Sj).

Some remarks may be useful to understand previous statement. i) says that large classes of common sets(like open or closed sets) are measurable. This is a good news, because most of sets we consider are openor closed. Just think to the case of sets like

O ={(x1, . . . , xn) ∈ Rn : gj(x1, . . . , xn) > 0, j = 1, . . . , k

}(open if gj ∈ C (Rn), j = 1, . . . , k),

C ={(x1, . . . , xn) ∈ Rn : gj(x1, . . . , xn) > 0, j = 1, . . . , k

}(closed if gj ∈ C (Rn), j = 1, . . . , k)

Furthermore, if our set S differs by an open/closed set by a measure 0 set, then S is measurable. Measurezero (or null) sets are particularly important. How can we imagine these sets? Here some examples:

61

• singletons S = {x∗} are measure 0 sets: indeed, if x∗ = (x∗1, . . . , x∗n) we may see

{x∗} = [x∗1, x∗1] × · · · × [x

∗n, x∗n], =⇒ λn({x∗}) = (x∗1 − x∗1) · · · (x

∗n − x∗n) = 0.

• countable sets S = {x∗j : j ∈ N} (like naturals or rationals in reals) are null sets: indeed, bycountable additivity

λn(S) =∑j

λn({x∗j }) =∑j

0 = 0.

• "lower dimensional" sets are in general measure zero sets: imagine a segment in the plane, sayfor simplicity S = {(x, c) : x ∈ [a, b]}. In R2 this is a null set: indeed

λ2(S) = λ2([a, b] × [c, c]) = (b − a)(c − c) = 0.To give a more precise and general statement we haveif g ∈ C (Rn) is non constant, =⇒ {(x1, . . . , xn) : g(x1, . . . , xn) = 0} is a null set.

Let f : E ⊂ Rn −→ [0,+∞[. We call trapezoid delimited by f the setTrap( f ) :=

{(x, y) ∈ Rn+1 : x ∈ E, 0 6 y 6 f (x)

}.

Definition 5.1.2. Let f : E −→ [0,+∞[. If Trap( f ) ∈Mn+1 we pose

(5.1.2)∫E

f := λn+1 (Trap( f )) .

First question: under which conditions is Trap( f ) is measurable? Here some important cases:

Proposition 5.1.3. Let f : E −→ [0,+∞[ be continuous on E closed or open. Then Trap( f ) ismeasurable. Thus, in particular, ∫

E

f

is well defined.Proof — For E closed it is easy to prove that Trap( f ) is closed too. For E open, Trap( f ) is not open in general.However, one can prove that

S := {(x, y) ∈ Rn+1 : x ∈ E, 0 < y < f (x)} is open,thus S ∈Mn. Clearly S ⊂ Trap( f ) (thus S\Trap( f ) = ∅) while

Trap( f )\S ={(x, y) ∈ Rn+1 : y = 0 ∨ y = f (x)

}.

Now, because{(x, y) ∈ Rn+1 : y = 0}, {(x, y) ∈ Rn+1 : y = f (x)} = {(x, y) ∈ Rn+1 : y − f (x) = 0}

are both defined by equations involving continuous functions (g(x, y) = y in the first case, g(x, y) = y − f (x) in thesecond), they have λn+1 = 0. We conclude that λn+1(Trap( f )\S) = 0 because Trap( f )\S is made of two measurezero sets and λn+1 is additive. The conclusion is that Trap( f ) differs by S open for a measure zero set, thus it ismeasurable.

Of course, we might need to consider discontinuous functions, but for purposes of this course f ∈ C (E)on E open or closed is more than sufficient. Throughout this Chapter we will develop a number ofmethods to compute integrals. These can be used to compute measures:

62

Proposition 5.1.4. If E is open or closed,

(5.1.3) λn(E) =∫E

1.

Proof — Just notice that if f ≡ 1,

Trap( f ) ={(x, y) ∈ Rn+1 : x ∈ E, 0 6 y 6 1

}≡ E × [0,1].

Thus, ∫E

1 dx = λn+1 (E × [0,1])f actor .= λn(E)λ1([0,1]) = λn(E).

So far, we defined the integral of a positive function. Let’s now consider f : E −→ R and define

f+(x) :=

f (x), if f (x) > 0,

0, if f (x) < 0,f−(x) :=

− f (x), if f (x) 6 0,

0, if f (x) > 0.Functions f± are called, respectively, positive part and negative part of f . Easily it turns out that

f ∈ C (E), =⇒ f± ∈ C (E).

Moreover, both f± are positive and, finally,f = f+ − f−, | f | = f+ + f−.

Notice that, if f ∈ C (E), E open/closed,∫E

| f | < +∞, ⇐⇒∫E

f+ < +∞,∫E

f− < +∞.

This justifies the following

Definition 5.1.5. Let f ∈ C (E), E open or closed in Rn. We say that f is integrable if∫E

| f | < +∞.

We pose ∫E

f :=∫E

f+ −∫E

f−.

and we denote the set of integrable functions with L1(E).

The properties of the integral are very similar to those of one dimensional integral:

Proposition 5.1.6. The following properties holds:i) (linearity) if f ,g ∈ L1(E) then α f + βg ∈ L1(E) for any α, β ∈ R and∫

E

(α f + βg) = α∫E

f + β∫E

g;

ii) (isotonicity) if f 6 g on E with f ,g ∈ L1(E) then∫E

f 6∫Eg;

iii) (triangular inequality) if f ∈ L1(E) then��∫E

f�� 6 ∫

E| f |;

63

iv) (decomposition) if f ∈ L1(E) and E = A ∪ B with A,B ∈Mn,∫E

f =∫A

f +∫B

f .

It remains now to develop an efficient method of calculus for integrals. This is based on two fundamentaltools: reduction formula and change of variables formula.

5.2. Reduction formula

In this Section we introduce the technique based on the reduction formula that allows to reduce thecalculus of a multiple variables integral to iterated one variable integrals. For pedagogical reasons wepresent first the case of double integrals, then we will extend to the general case.

5.2.1. Double Integrals. To understand the idea, let’s consider the problem to compute∫E

f (x, y) dxdy.

As we know, integrals are continuous versions of discrete sums. Thinking to these we could write∑(x,y)∈E

f (x, y) =∑x∈R

©«∑

y∈R : (x,y)∈Ef (x, y)ª®¬ =

∑y∈R

©«∑

x∈R : (x,y)∈Ef (x, y)ª®¬ .

If the sums were finite, there wouldn’t be any problem in reordering terms and summing as we prefer,so the previous formula would be a consequence of associativity and commutativity. However, whensum contains infinitely many terms the story is much more complicate. Leaving aside for a moment thisproblem, we might expect that the analogous for integrals of the previous formula is∫

E

f (x, y) dxdy =∫R

(∫Ey

f (x, y) dy)

dx =∫R

(∫Ex

f (x, y) dx)

dy.

The two sets Ex and Ey are defined as

Ex := {y ∈ R : (x, y) ∈ E} , (x − section), Ey := {x ∈ R : (x, y) ∈ E} , (y − section).

Notice that, fixed x ∈ R, Ex is the set of ordinates y of points of E with abscissa x, that is y such that(x, y) ∈ E . In other words, Ex is the projection on the y−axis of the "slice" of E along the vertical straightline at abscissa x. It turns out that if f ∈ L1 the reduction formula holds true.

E

Ex

xX

Y

E

Ey

X

y

Y

64

Proposition 5.2.1. Let f ∈ L1(E), then

(5.2.1)∫E

f (x, y) dxdy =∫R

(∫Ex

f (x, y) dy)

dx =∫R

(∫Ey

f (x, y) dx)

dy.

Remark 5.2.2. Notice that Ex (and Ey) may be empty for certain values of x (resp y). For such x (y),clearly

∫Ex

f = 0 (∫Ey f = 0). Therefore∫

R

(∫Ex

f (x, y) dy)

dx =∫x∈R : Ex,∅

∫Ex

f (x, y) dy dx.

However, for future use we prefer to keep a lighter notation as in (5.2.1).

The (5.2.1) requires f ∈ L1(E), that is ∫E

| f (x, y)| dxdy < +∞.

To check this, in principle one should compute a double integral. Notice that if we know f ∈ L1(E) then,by the reduction formula,∫

E

| f (x, y)| dxdy =∫R

(∫Ex

| f (x, y)| dy)

dx =∫R

(∫Ey| f (x, y)| dx

)dy,

so in particular ∫R

(∫Ex

| f (x, y)| dy)

dx,∫R

(∫Ey| f (x, y)| dx

)dy < +∞.

It turns out that also the vice versa holds true:Proposition 5.2.3. Let f ∈ C (E), E ⊂ R2 open/closed set. If one of the following iterated integrals

(5.2.2)∫R

(∫Ex

| f (x, y)| dy)

dx,∫R

(∫Ey| f (x, y)| dx

)dy

is finite, then f ∈ L1(E) and reduction formula (5.2.1) holds.Combining the previous Propositions we have an algorithm to check if f ∈ L1(E) and to compute itsintegral by using the reduction formula: first, one check if one of the (5.2.2) is finite (which one of thetwo is indifferent and the choice could be done in terms of computational ease); second, one uses (5.2.1)to compute the integral. Notice that, in particular, if f > 0 the check (5.2.2) leads at same time to thecalculation of the integral by (5.2.1).

Example 5.2.4. Discuss if f (x, y) := x3e−yx2∈ L1([0,+∞[×[1,2]) and compute its integral.

Sol. — Clearly f ∈ C (E) where E = [0,+∞[×[1,2] is closed. Applying (5.2.2), notice that if E = [0,+∞[×[1,2],Ex = ∅ if x < 0, Ex = [1,2] if x > 0, therefore∫

R

∫Ex

| f | dy dx =

∫ +∞0

(∫ 2

1x3e−yx

2dy

)dx =

∫ +∞0

x[−e−yx

2]y=2

y=1dx

=

∫ +∞0

xe−x2− xe−2x2

dx =

[−e−x

2

2

] x=+∞x=0

−

[−e−2x2

4

] x=+∞x=0

=14.

65

We deduce f ∈ L1 and because f > 0, thus | f | = f , the same calculation and (5.2.1) gives∫[0,+∞[×[1,2] f = 1

4 .

Example 5.2.5. Discuss if f (x, y) := e−x ∈ L1(E) where E = {(x, y) ∈ R2 : x > 0, 0 6 y 6 x2}. Insuch case compute the integral of f on E .Sol. — Clearly f ∈ C (E) where E is closed (defined by large inequalities on continuous functions). Applying(5.2.2), notice that Ex = ∅ if x < 0, Ex = [0, x2] if x > 0, therefore∫

R

∫Ex

| f | dy dx =

∫ +∞0

(∫ x2

0e−x dy

)dx =

∫ +∞0

x2e−x dx =∫ +∞

0x2(−e−x)′ dx

=[−x2e−x

] x=+∞x=0 +

∫ +∞0

2xe−x dx = 2∫ +∞

0x(−e−x)′ dx

= 2[[−xe−x]x=+∞x=0 +

∫ +∞0

e−x dx]= 2 [−e−x]x=+∞x=0 = 2.

Therefore f ∈ L1(E) and because f > 0 the same calculation and (5.2.1) gives∫E

f = 2.

A remarkable example of (5.2.1) is obtained by taking f ≡ 1. Recalling that∫E

1 = λ2(E) we obtain

(5.2.3) λ2(E) =∫R

(∫Ex

1 dy)

dx =∫Rλ1(Ex) dx =

∫Rλ1(Ey) dy.

Example 5.2.6. Compute the area of a disk of radius r .Sol. — Because the measure is translations–invariant, we can center the disk into the origin. So, let’s consider

E ={(x, y) ∈ R2 : x2 + y2 6 r2} .

This set E is closed, hence measurable. Applying the integration by slices we have

λ2(E) =∫Rλ1(Ex) dx.

Let’s determine an x−section. We have

y ∈ Ex, ⇐⇒ (x, y) ∈ E, ⇐⇒ x2 + y2 6 r2, ⇐⇒ y2 6 r2 − x2, ⇐⇒ y ∈[−√

r2 − x2,√

r2 − x2].

Of course we need r2 − x2 > 0, that is x2 6 r2, |x | 6 r , otherwise Ex = ∅. Therefore

λ2(E) =∫Rλ1(Ex) dx =

∫|x |6r

λ1

( [−√

r2 − x2,√

r2 − x2] )

dx =∫|x |6r

2√

r2 − x2 dx.

Being x 7−→√

r2 − x2 continuous on [−r,r], the last integral is equal to a Riemann one, so

λ2(E) =∫ r

−r

2√

r2 − x2 dx = 4∫ r

0

√r2 − x2 dx = 4r

∫ r

0

√1 −

x2

r2 dx.

Therefore, setting xr = sin θ, θ ∈ [0, π2 ],

λ2(E) = 4r∫ π/2

0

√1 − (sin θ)2 r cos θ dθ = 4r2

∫ π/2

0(cos θ)2 dθ.

66

Now∫(cos θ)2 =

∫cos θ(sin θ)′ = cos θ sin θ +

∫(sin θ)2 = 1

2 sin(2θ) + θ −∫(cos θ)2 hence

λ2(E) = 4r2[14

sin(2θ) +θ

2

]θ=π/2θ=0

= πr2.

Warning! If f < L1 the reduction formula might be false even if the iterated integral are finite.

Example 5.2.7. Letf (x, y) =

x − y

(x + y)3, (x, y) ∈ E := [0,1]2.

Then∫R

(∫Ex

f dy)

dx ,∫R

(∫Ey f dx

)dy. Hence, in particular, f < L1([0,1]2).

Sol. — Notice first that

Ex = {y ∈ R : (x, y) ∈ [0,1]2} =∅, x < [0,1],

[0,1] x ∈ [0,1]

and similarly for Ey . Therefore

∫Ey

f (x, y) dx =

0, y < [0,1],∫ 1

0

x − y

(x + y)3dx =

∫ 1

0

1(x + y)2

dx − 2y∫ 1

0

1(x + y)3

dx. y ∈ [0,1].

Except for y = 0 (therefore for a measure 0 set) both integrals are finite and their value is[(x + y)−1

−1

] x=1

x=0− 2y

[(x + y)−2

−2

] x=1

x=0=

1y−

1y + 1

+ y

(1

(y + 1)2−

1y2

)= −

1(y + 1)2

.

Hence ∫R

(∫Ey

f (x, y) dx)

dy =∫ 1

0

(−

1(y + 1)2

)dy =

[(y + 1)−1]y=1

y=0 =12− 1 = −

12.

Exchanging x with y we obtain the same result except for the sign:∫R

(∫Ex

f (x, y)dy)

dx = 12 .

5.2.2. General Multiple Integrals. The previous mechanism can be extended to functions f of nvariables. Let f = f (z1, . . . , zn) and imagine we group (z1, . . . , zn) into two blocks, one of k variablesand the remaining of n − k variables. For simplicity with notations we write

f = f (x, y), where x = (x1, . . . , xk) ∈ Rk, y = (xk+1, . . . , xn) ∈ Rn−k .

As above, we will denote by Ex (resp. Ey) the x−section (resp. y−section) of E defined as

Ex := {y ∈ Rn : (x, y) ∈ E} , Ey := {x ∈ Rm : (x, y) ∈ E} .

Be careful because now Ex ⊂ Rn while Ey ⊂ Rm. With these notations we have the

67

Theorem 5.2.8 (Fubini–Tonelli). Let f ∈ L1(E), E ⊂ Rm+n. Then the reduction formula holds

(5.2.4)∫E

f =∫Rm

(∫Ex

f (x, y) dy)

dx =∫Rn

(∫Ey

f (x, y) dx)

dy.

Moreover, if f ∈ C (E) and one among∫Rm

(∫Ex

| f (x, y)| dy)

dx,∫Rn

(∫Ey| f (x, y)| dx

)dy,

is finite, then f ∈ L1(E) (and the reduction formula (5.2.4) holds). In particular, by taking f = 1 wehave the slicing formula

(5.2.5) λm+n(E) =∫Rm

λn(Ex) dx =∫Rnλm(Ey) dy.

Fubini–Tonelli theorem is a versatile tool to integrate functions of several variables. For instance: considera function of three variables f = f (x, y, z) ∈ C (E), E ⊂ R3 open/closed. In this common case, thethree variables may be grouped is six different ways, this leading to six different possible applications ofreduction formula:

x and (y, z),∫E

f =∫R

(∫(y,z)∈Ex

f dydz)

dx =∫R2

(∫x∈E(y ,z)

f dx)

dydz,

y and (x, z),∫E

f =∫R

(∫(x,z)∈Ey

f dxdz)

dy =∫R2

(∫y∈E(x ,z)

f dy)

dxdz,

z and (x, y),∫E

f =∫R

(∫(y,z)∈Ex

f dydz)

dx =∫R2

(∫x∈E(y ,z)

f dx)

dydz,

Which choice is the best one depends by the complexity of calculus.

Example 5.2.9. Compute the volume of a rugby ball E ={(x, y, z) ∈ R3 : x2+y2

a2 +z2

b2 6 1}, (a, b > 0).

Sol. — Clearly E is closed in R3, hence measurable. Slicing E along the z-axis,

m3(E) =∫R

m2(Ez) dz.

Now,

(x, y, z) ∈ E, ⇐⇒x2 + y2

a2 6 1 −z2

b2 , ⇐⇒ (x, y) ∈ B

(02,

√1 −

z2

b2

]=: Ez

68

This of course if 1 − z2

b2 > 0, that is z2 6 b2, namely |z | 6 b, otherwise Ez = ∅. It follows that

m3(E) =∫|z |6b

m2

(B

(02,a

√1 −

z2

b2

])dz =

∫|z |6b

πa2(1 −

z2

b2

)dz R=L=

∫ b

−b

πa2(1 −

z2

b2

)dz

= πa2

([z]b−b −

[z3

3b2

]b−b

)= πa2

(2b −

23

b)= π

43

a2b.

Taking a = b = r we obtain the volume of a sphere of radius r , the well known 43πr3.

5.3. Change of variables

Let T : Rn −→ Rn a transformation. If T is a linear bijection we know that

λn(T(E)) = | det T |λn(E).

What happens if T is a general (non linear) bijection?

Theorem 5.3.1. Let T : E ⊂ Rn −→ T(E) be a diffeomorphism (that is T,T−1 ∈ C 1) on E open/closedset. Then

(5.3.1) λn(T(E)) =∫E

| det T ′(ξ)| dξ.

Proof — (sketch): We decompose E as disjoint union of neighborhoods of some of its points, let’s say

E =⋃j

Ux j .

Then, by countable additivity,λn(T(E)) =

∑j

λn(T(Ux j )).

If T is regular (differentiable) then T(x) = T(x0)+T ′(x0)(x − x0)+ o(x − x0) ∼x0 T(x0)+T ′(x0)(x − x0). The senseof ∼x0 is that T(x) can be replaced by T(x0)+T ′(x0)(x − x0) in a neighborhood of x0 and the approximation is moreprecise smaller is this neighborhood. Therefore, we may expect that

λn(T(Ux j )) ≈ λn(T(xj) + T ′(xj)(Ux j − xj)) = λn(T ′(xj)(Ux j − xj)) = | det T ′(xj)|λn(Ux j − xj)

= | det T ′(xj)|λn(Ux j ).

Hence

λn(T(E)) ≈∑j

| det T ′(xj)|λn(Ux j ) ≈

∫E

| det T ′(ξ)| dξ.

Of course this is not yet a proof and it needs a lot of technical work to make it a rigorous argument, but this is themain idea on which is based.

69

We may review (5.3.1) as∫T (E)

dx =∫E

| det T ′(ξ)| dξ,Φ:=T−1

⇐⇒

∫F

dx =∫Φ(F)

| det(Φ−1)′(ξ)| dξ.

This last is a special case of change of variable formula:

Theorem 5.3.2 (change of variables formula). Let f : F −→ R, f ∈ C (F), F open/closed, andΦ : F −→ Φ(F) be a diffeomorphism, that is Φ,Φ−1 ∈ C 1. Then

(5.3.2)∫F

f (x) dxy=Φ(x), x=Φ−1(y)

=

∫Φ(F)

f (Φ−1(ξ))| det(Φ−1)′(ξ)| dξ.

Proof — (sketch) As in the previous proof, divide F in a disjoint union of small neighbourhoods,

F =⋃j

Ux j ,

in such a way that ∫F

f =∑j

∫Ux j

f .

We may choose Ux j in such a way that f (x) ≈ f (xj) for all x ∈ Ux j (by continuity). Thus∫Ux j

f ≈∫Ux j

f (xj) = f (xj)∫Ux j

= f (xj)∫Φ(Ux j

)

| det(Φ−1)′ |.

Now, because Φ is a diffeomorphism, we may imagine that Φ(Ux j ) = VΦ(x j ) is a neighbourhood of Φ(xj). Thus, ifξ ∈ VΦ(x j ), by continuity f (Φ−1(ξ)) ≈ f (Φ−1(Φ(xj))) = f (xj) for all ξ ∈ VΦ(x j ). Furthermore,⋃

j

VΦ(x j ) =⋃j

Φ(Ux j ) = Φ

(⋃j

Ux j

)= Φ(F).

Summing up all previous remarks we would have∫F

f ≈∑j

f (xj)∫Φ(Ux j

)

| det(Φ−1)′ | =∑j

∫VΦ(x j )

f (Φ−1(ξ))| det(Φ−1)′(ξ)| =

∫Φ(F)

f (Φ−1(ξ))| det(Φ−1)′(ξ)| dξ.

Example 5.3.3. Compute∫16xy62, 0<ax6y6 x

a

y4 arctan(xy)(x2 + y2)2

dxdy, 0 < a < 1.

Sol. — The domain is closed in R2, hence measurable and f ∈ C .

a 2 a1

a

1

a�2

x

y

70

Notice thaty4 arctan(xy)(x2 + y2)2

=( y

x

)4 arctan(xy)(1 +

( yx

)2)2 .

It seems therefore natural to introduce the new variables

ξ = xy, η =y

x, (ξ, η) := Ψ(x, y),

where Ψ :]0,+∞[2−→]0,+∞[2, Ψ(x, y) = (xy, yx ) is clearly C 1. We need Ψ−1. If (ξ, η) ∈]0,+∞[2 thenξ = xy,

η =yx ,

⇐⇒

ξ = ηx2,

y = ηx,⇐⇒

x =

√ξη ,

y =√ξη,

⇐⇒ Ψ−1(ξ, η) =

(√ξ

η,√ξη

).

Therefore

I(a) :=∫

16xy62, 0<ax6y6 xa

y4 arctan(xy)(x2 + y2)2

dxdy =∫

16ξ62, a6η6 1a

η4

(1 + η2)2arctan ξ | det(Ψ−1)′(ξ, η)| dξdη.

and because| det(Ψ−1)′(ξ, η)| =

1| detΨ′(Ψ−1(ξ, η))|

,

with

Ψ′(x, y) =

y x

−y

x21x

, =⇒ detΨ′(x, y) =y

x+ x

y

x2 = 2y

x= 2η.

we have

I(a) =∫

16ξ62, a6η6 1a

η4

(1 + η2)2arctan ξ

12η

dξdη =12

(∫ 2

1arctan ξ dξ

) (∫ 1a

a

η3

(1 + η2)2dη

).

Now ∫ 2

1arctan ξ dξ = [ξ arctan ξ]21 −

∫ 2

1

ξ

1 + ξ2 dξ = 2 arctan 2 −π

4−

12

log52,

while ∫ 1a

a

η3

(1 + η2)2dη =

∫ 1a

a

η

1 + η2 dη −∫ 1

a

a

η

(1 + η2)2dη = − log a +

12

1 − a2

1 + a2 .

5.3.1. Polar coordinates in R2. A very important change of variable in plane integration isx = ρ cos θ,

y = ρ sin θ,⇐⇒ (x, y) = Ψ(ρ, θ) = (ρ cos θ, ρ sin θ).

Here we may notice that change of variable is defined in the form (x, y) = Ψ(ρ, θ). This means that,referring to notations of (5.3.2), present Ψ is just Φ−1. Thus

det(Φ−1)′ = detΨ′ = det

cos θ −ρ sin θ

sin θ ρ cos θ

= ρ(cos2 θ + sin2 θ) = ρ,

71

and (5.3.2) becomes

(5.3.3)∫E

f (x, y) dxdy =∫Epol

f (ρ cos θ, ρ sin θ)ρ dρdθ,

where Epol is E in polar coordinates.

Example 5.3.4. Compute ∫R2

e−√

x2+y2dxdy.

Sol. — We have∫R2

e−√

x2+y2dxdy =

∫ρ>0,θ∈[0,2π]

e−ρρ dρdθ =∫ +∞

0

(∫ 2π

0e−ρρ dθ

)dρ = 2π

∫ +∞0

ρe−ρ dρ

= 2π([−ρe−ρ]ρ=+∞

ρ=0 +

∫ +∞0

e−ρ dρ)= 2π.

Example 5.3.5 (Gaussian integral). A very beautiful (and relevant) application of the (5.3.3) is theformula ∫

Re−

x22 dx =

√2π.

More in general: if C is a d × d positive symmetric matrix,

(5.3.4)∫Rd

e−12C−1x ·x dx =

√(2π)d det C.

Sol. — Let’s start by the integral∫R2

e−x2+y2

2 dxdy =∫R

(∫R

e−x2+y2

2 dx)

dy =∫R

e−x22

(∫R

e−y22 dx

)dy =

(∫R

e−x22 dx

)2.

On the other hand, by (5.3.3)∫R2

e−x2+y2

2 dxdy =∫ +∞

0

(∫ 2π

0e−

ρ22 ρ dθ

)dρ = 2π

∫ +∞0

e−ρ22 ρ dρ = 2π

[e−

ρ22

]ρ=+∞ρ=0

= 2π,

and by this the conclusion follows.To compute (5.3.4) notice first that, being C symmetric, it is diagonalizable: this means that there exists T

invertible such that T−1CT = diag(σ1, . . . ,σd). Furthermore, because C is symmetric, T is also orthogonal, that isT−1 = T t (transposed matrix). Therefore C = T DT−1, hence∫

Rde−

12C−1x ·x dx =

∫Rd

e−12 (TDT 1−)−1x ·x dx =

∫Rd

e−12 (TD−1T 1−)x ·x dx =

∫Rd

e−12 D−1T−1x ·T−1x dx.

Now, set y = T−1x, in such a way that x = T y and∫Rd

e−12 D−1T t x ·T t x dx =

∫Rd

e−12 D−1y ·y | det T | dy =

∫Rd

e−12 D−1y ·y dy.

72

Last = is justified because, being T orthogonal, TT t = I, hence 1 = det(TT t ) = det T det T t = (det T)2 by which| det T | = 1. Moreover,

D1−y · y =∑j

1σj

y2j ,

therefore∫Rd

e−12 D−1y ·y dy =

∫Rd

d∏j=1

e−

y2j

2σ j dyj =d∏j=1

∫R

e−

y2j

2σ j dyjx j=

yj√σ j

=

d∏j=1

√σj

∫R

e−x22 dx =

√(2π)dσ1 · · ·σd .

To conclude just notice that

σ1 · · ·σd = det D = det(T−1CT) = det T−1 det C det T = det C.

5.3.2. Spherical and cylindrical coordinates. The analogous of polar coordinates for functions ofthree variables are spherical coordinates:

x = ρ cos θ sin ϕ,y = ρ sin θ sin ϕ,z = ρ cos ϕ.

(ρ, θ, ϕ) ∈ [0,+∞[×[0,2π] × [0, π].

Also in this case the change of variable is defined in the form

(x, y, z) = Ψ(ρ, θ, ϕ),

thus, referring to (5.3.2), Ψ = Φ−1. Hence,

det(Φ−1)′ = det

cos θ sin ϕ −ρ sin θ sin ϕ ρ cos θ cos ϕsin θ sin ϕ ρ cos θ sin ϕ ρ sin θ cos ϕ

cos ϕ 0 −ρ sin ϕ

= ρ2 sin ϕ.

Therefore, (5.3.2) reads as∫E

f (x, y, z) dxdydz =∫Espher

f (ρ cos θ sin ϕ, ρ sin θ sin ϕ, ρ cos ϕ)ρ2 sin ϕ dϕ dθ dρ.

Here Espher is E in spherical coordinates. This type of change of variable is often useful when f hassome spherical symmetry, that is it depends on x2 + y2 + z2.

Example 5.3.6. Using spherical coordinates, compute the volume of a sphere of radius r .Sol. —We have

λ3

({x2 + y2 + z2 6 r2}

)=

∫x2+y2+z26r2

dxdydz =∫

06ρ6r , 06θ62π, 06ϕ6πρ2 sin ϕ dρdθdϕ

= 2π(∫ π

0sin ϕ dϕ

) (∫ r

0ρ2 dρ

)=

43πr3.

73

When f has not a central symmetry but it is symmetric respect to some of the axes, a further variant ofpolar coordinates may be useful. Let first introduce this system of coordinates defined as

x = ρ cos θ,y = ρ sin θ,z = z.

(ρ, θ, z) ∈ [0,+∞[×[0,2π] × R.

Also in this case the change of variables is defined in the form

(x, y, z) = Ψ(ρ, θ, z), where Ψ = Φ−1.

Being,

detΨ′ = det

cos θ −ρ sin θ 0sin θ ρ cos θ 0

0 0 1

= ρ,according to (5.3.2) we have∫

E

f (x, y, z) dxdydz =∫Ecil

f (ρ cos θ, ρ sin θ, z)ρ dρdθdz.

This change of variables is particularly useful in the case of functions symmetric respect to the z axis(that is depending on x2 + y2 that becomes ρ2 in new coords).

Example 5.3.7. Compute the volume of the rugby ball E = {(x, y, z) ∈ R3 : x2+y2

a2 +z2

b2 6 1} by adaptingcylindrical coordinates.Proof — Adapting the cylindrical coords (x, y, z) = Ψ−1(ρ, θ, z) := (aρ cos θ,aρ sin θ, bz) we have

det(Ψ−1)′ = det

a cos θ −aρ sin θ 0a sin θ aρ cos θ 0

0 0 b

= ba2ρ,

thereforem3(E) =

∫ρ2+z261, ρ>0, θ∈[0,2π], z∈R

ba2ρ dρdθdz = 2πa2b∫ρ2+z261, ρ>0

ρ dρdz.

To compute the last integral we may use polar coords for (ρ, z) = (r cosα,r sinα). Then∫ρ2+z261, ρ>0

ρ dρdz =∫− π2 6α6

π2 , 06r61

(r cosα)r drdα =∫ π

2

− π2

cosα dα∫ 1

0r2 dr =

23.

Moral: m3(E) = 4π3 a2b.

5.4. Barycenter, center of mass, inertia moments

Through multiple integrals we can define several quantities relevant in Geometry and Physics. To fixideas consider a set E ⊂ R3. We call barycenter of E the point (x, y, z) defined as

x =1

λ3(E)

∫E

x dx dy dz, y =1

λ3(E)

∫E

y dx dy dz, z =1

λ3(E)

∫E

z dx dy dz.

In other words, the barycenter is the point whose coords are the mean values of the coords of E . Withspecial symmetries some of the coords of the barycenter may vanish. For instance, if E is symmetric with

74

respect to the plane yz, that is (x, y, z) ∈ E iff (−x, y, z) ∈ E), then x = 0. Indeed, if Φ(x, y, z) = (−x, y, z)we have Φ(E) = E therefore, by change of variables,∫

E

x dxdydz =∫Φ(E)

x dxdydz =∫E

(−x)| detΦ′(x, y, z)| dxdydz = −∫E

x dxdydz

by which∫E

x dxdydz = 0.If E represents a solid body with density of mass % = %(x, y, z), the total mass is, by definition,

µ(E) :=∫E

%(x, y, z) dxdydz.

In Physics it is then important the center of mass: it is the point where the sum of all the forces actingon E could be applied to get the same effect. This point has coords (xG, yG, zG)

xG =1

µ(E)

∫E

x%(x, y, z) dxdydz, yG =1

µ(E)

∫E

y%(x, y, z) dxdydz, zG =1

µ(E)

∫E

z%(x, y, z) dxdydz.

If the body is homogeneous (that is % ≡ %0 ∈ R) the center of mass coincide with the barycenter as it iseasy to see.

Another important quantity for Physics is the inertia moment with respect to some axis. Forinstance, if the axis is the z one, this is defined by

Iz :=∫E

(x2 + y2)%(x, y, z) dxdydz.

Example 5.4.1. Determine the barycenter of a spherical cap E := {(x, y, z) : x2 + y2 + z2 6 r2, z > h}con 0 6 h < r .Sol. — By symmetries, it is evident that x = y = 0. Let’s compute

z =1

λ3(E)

∫E

z dxdydz.

It seems convenient to slice E perpendicularly to the z−axis:

λ3(E) =∫ r

h

(∫x2+y26r2−z2

dxdy)

dh =∫ r

h

π(r2 − z2) dz = πr2(r − h) − π[

z3

3

]z=rz=h

= π(r − h)(r2 −

13(r2 + rh + h2)

).

Similarly∫E

z dxdydz =

∫ r

h

(∫x2+y26r2−z2

z dxdy)

dz =∫ r

h

z(∫

x2+y26r2−z2dxdy

)dz =

∫ r

h

zπ(r2 − z2) dz

= πr2[

z2

2

]z=rz=h

− π

[z4

4

]z=rz=h

= πr2 r2 − h2

2− π

r4 − h4

4= π

r2 − h2

2

(r2 −

r2 + h2

2

)= π(r2 − h2)2

4.

By this we get z. In the case h = 0 (that is when E is the half-sphere) we have z = 38r .

75

Let D ⊂ R3 by a domain obtained by a rotation around one of the axes of a plane set E . To fix ideas,let’s assume that the rotation be around the z−axis of a domain E in the plane yz. This domain can beidentified by {(0, y, z) : (y, z) ∈ E} ⊂ R3. Therefore, D can be represented as

D = {(y cos θ, y sin θ, z) : (y, z) ∈ E, θ ∈ [0,2π]} = Φ(E × [0,2π]),where Φ is nothing but the cylindrical coords map.

E

y

z

By the formula of change of variables

λ3(D) =∫E×[0,2π]

JΦ(y, θ, z) dy dθ dz =∫E×[0,2π]

y dy dθ dz = 2π∫E

y dydz

that gives the Pappo’s Theorem:(5.4.1) λ3(D) = 2πλ2(E)y.

Example 5.4.2. Let’s compute the volumeof a thorusTr ,R :={(x, y, z) ∈ R3 : (

√x2 + y2 − R)2 + z2 6 r2

}(0 < r < R).Sol. — By (5.4.1)

λ3(Tr ,R) = 2πm2

({(y − R)2 + z2 6 r2

)y = 2π1πr2y = 4π2r2y.

Here y it’s the ordinate of the barycenter of the disk E := {(y − R)2 + z2 6 r2}, so

y =1

λ2(E)

∫E

y dydz =1πr2

∫(y−R)2+z26r2

y dydz.

Changing to polar coord y − R = ρ cos θ, z = ρ sin θ, we have easily

y =1πr2

∫ 2π

0

(∫ r

0ρ(R + ρ cos θ) dρ

)dθ =

1πr2 2π

r2

2R = R,

(as it is natural!). Hence λ3(Tr ,R) = 4π2r2R.

5.5. ExercisesExercise 5.5.1. Compute

1.∫

06y61, 06x61−y2 xey dxdy. 2.∫

06y61−x2x

2+y dxdy. 3.∫|y |61−x2

11+y dxdy

4.∫

06x,y61, 06z63−x+y x sin(πy) dxdydz4. 5.∫x>0, y>0, x+y+z61 xyz dxdydz.

6.∫

06x61, 06y62, 06z66−x2−y2 x log(1 + y) dxdydz.

76

Exercise 5.5.2. Compute:

1.∫[0,1]2

emax{x2 ,y2 } dxdy. 2.∫[0,1]×[2,4]

1(x − y)2

dxdy. 3.∫[0,+∞[×[1,+∞[

e−xy4

dxdy.

4.∫

16x62, 1x 6y6x

xy

dxdy. 5.∫[0,1]3

emax{x,y,z } dxdy. 6.∫[1,+∞[3

y3z8e−xy2z3

dxdydz.

Exercise 5.5.3. Compute

1.∫D

x√y2 − x2 dxdy, D = {(x, y) ∈ R2 : 0 6 y 6 1, 0 6 x 6 y}.

2.∫D

x2e−x2

1 + (xy)2dxdy, D = {(x, y) ∈ R2 : |xy | 6 1}.

3.∫D

zy2√

x2 + zy dxdydz, D = {(x, y, z) ∈ R3 : 0 6 z 6 x2, 0 6 x 6 1, 0 6 y 6 1}.

Exercise 5.5.4. For which values α ∈ R the function fα(x, y) := 1(x−y)α belongs to L1([1,+∞[×[0,1])? In this case

compute the integral∫[1,+∞[×[0,1] fα.

Exercise 5.5.5. Let D := {(x, y) ∈ R2 : x > 0, y > 0, x2 + y2 6 r2}. Draw D and describe it in polar coords.Determine its barycenter and compute the integral

∫D

x + y

x2 + y2 dxdy.

Exercise 5.5.6 (polar, spherical, cylindrical coords). Draw (if possible) and compute the volume of

1.{(x, y, z) : 9(1 −

√x2 + y2)2 + 4z2 6 1

}. 2.

{(x, y, z) : x2 + y2 + z2 6 r2,

(x − r

2)2+ y2 6 r2

4

}.

3.{(x, y, z) ∈ R3 : x2

a2 +y2

b2 +z2

c2 6 1}, (a, b, c > 0). 4.

{(x, y, z) : x2 + y2 6 1, x2 + z2 6 1, y2 + z2 6 1

}.

5.{(x, y, z) ∈ R3 : x2 + y2 6 4, 4x2 + 4y2 + z2 6 64

}. 6.

{(x, y, z) ∈ R3 : x2 + y2 + z2 6 16, x2 + y2 > 4

}7.

{(x, y, z) ∈ R3 : z >

√x2 + y2, x2 + y2 + z2 6 1

}. 8.

{(x, y, z) ∈ R3 : z > x2 + y2, z 6 18 − x2 − y2}.

77

Exercise 5.5.7. Discuss existence and (if possible) the value of the following integrals:

1.∫R2

1cosh(x2 + y2)

dxdy. 2.∫x2+y264

√4 − x2 − y2 dxdy

3.∫x2+2y261

11 + x2 + 2y2 dxdy. 4.

∫R2

11 + (x2 + 2y2)2

dxdy.

5.∫x2+y2616, −56z64

√x2 + y2 dxdydz. 6.

∫R3

√x2 + y2 + z2e−(x

2+y2+z2) dxdydz.

7.∫R3

11 + (x2 + 2y2 + 3z2)2

dxdydz. 8.∫R2

11 + (x2 + xy + y2)2

dxdy.

9.∫R3

e−(x2+y2+z2−xy+yz−xz) dxdydz. 10.

∫R3

11 + x4 + y4 + z4 dxdydz.

Exercise 5.5.8. By using carefully the suggested change of variables, compute

1.∫D

xy dxdy, D = {(x, y) ∈ R2 : 1 6 xy 6 3, x 6 y 6 3x}, (u = xy, v =y

x).

2.∫D

y2 dxdy, D = {(x, y) ∈ R2 : 1 6 xy 6 2, 1 6 xy2 6 2}. (u = xy, v = xy2).

3.∫D

√x2 − y2 dxdy, D = {(x, y) ∈ R2 : 1 6 x2 − y2 6 2, px 6 y 6 qx} (−1 < p < q < 1). (u = x2 − y2, v =

y

x).

Exercise 5.5.9. Let a > 1 and

Ea :={(x, y) ∈ R2 :

1ax6 y 6

1x, x2 6 y 6 ax2

}.

Draw Ea. Show that Φ(x, y) :=(xy, y

x2

)is a diffeomorphism modulo null sets on Ea. Use this to compute

I(a) :=∫Ea

x2

yexy dxdy.

Compute lima→+∞ I(a) and check if I =∫x26y6 1

x

x2

y exy dxdy.

Exercise 5.5.10 (?). Let D := {(x, y) ∈ R2 : 0 6 x 6 y} and

f (x, y) :=x3/2√y − x

e−(xy)3/2, (x, y) ∈ D.

Show that Φ(x, y) := (xy, x/y) is a diffeomorphism modulo null sets on D. Use this to say if f ∈ L1(D). In suchcase compute

∫D

f .

Exercise 5.5.11 (?). Let D := {(x, y) ∈ [0,+∞[2 : xy > 1}, and f (x, y) := log(xy)y(x+y2)7

, (x, y) ∈ D. Show that

Φ(x, y) := (xy, y2

x ) is a diffeomorphism modulo null sets on E . Use this to say if f ∈ L1(D). In such case compute∫D

f .

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